Page 1 of 2
RIGHT TRIANGLES
AND TRIGONOMETRY
How do builders find the dimensions
of a skywalk support beam?
524
Page 2 of 2
CHAPTER
9
APPLICATION: Support Beam
When constructing a skywalk
for a skyscraper, builders often
use mathematics to measure
some lengths indirectly.
One method for indirect measurement involves using relationships
in a right triangle.
B
A
C
support
beams
D
E
Think & Discuss
Use the diagram of a skywalk connecting two
buildings for Exercises 1 and 2.
1. Given the information in the diagram, how do
you know that the skyscrapers are parallel to
each other?
2. How do you know that the two right triangles
created by the skywalk, beams, and buildings
are congruent?
Learn More About It
INT
You will see how the Pythagorean Theorem is used
to find the length of a skywalk support beam in
Example 4 on p. 537.
NE
ER T
APPLICATION LINK Visit www.mcdougallittell.com
for more information about skyscrapers.
525
Page 1 of 1
Study Guide
CHAPTER
9
PREVIEW
What’s the chapter about?
Chapter 9 is about right triangles and a related branch of mathematics called
trigonometry. In Chapter 9, you’ll learn
•
about properties related to general right triangles, similar right triangles, and
special right triangles.
•
about some applications of right triangles, including trigonometry, or triangle
measurement, and vectors.
K E Y VO C A B U L A RY
PREPARE
Review
• similar polygons, p. 473
• converse, p. 72
• right triangle, p. 194
• altitude, p. 281
• proportion, p. 459
• geometric mean, p. 466
New
• Pythagorean triple, p. 536
• special right triangles,
p. 551
• trigonometric ratio, p. 558
• angle of elevation, p. 561
• solve a right triangle, p. 567
• magnitude of a vector,
p. 573
• direction of a vector, p. 574
• sum of two vectors, p. 575
Are you ready for the chapter?
SKILL REVIEW Do these exercises to review key skills that you’ll apply in this
chapter. See the given reference page if there is something you don’t understand.
Exercises 1 and 2 refer to ¤JKL with m™J = 30° and m™K = 60°.
1. Find m™L. Classify ¤JKL as acute, right, or obtuse. (Review pp. 194–197)
2. Sketch ¤JKL. Label its legs, hypotenuse, and altitudes. (Review pp. 195, 281)
3. Draw a vector in a coordinate plane whose component form is ⟨5, º2⟩.
(Review p. 423)
x
x+3
4. Solve the proportion: = . (Review p. 459)
3
5
5. Refer to ¤JKL in Exercises 1 and 2 above. Sketch a triangle, ¤ABC, that is
similar to ¤JKL. Explain how you know the triangles are similar.
(Review p. 481)
STUDY
STRATEGY
Here’s a study
strategy!
What Do You Know?
Make a list of what you alread
y kn
triangles and trigonometry. Ma ow about right
ke another list of what
you expect to learn about the
se topics. After studying
Chapter 9, review these lists
as you review the
chapter to see what you have
learned.
526
Chapter 9
Page 1 of 8
9.1
Similar Right Triangles
What you should learn
GOAL 1 Solve problems
involving similar right triangles
formed by the altitude drawn
to the hypotenuse of a right
triangle.
Use a geometric
mean to solve problems, such
as estimating a climbing
distance in Ex. 32.
GOAL 2
GOAL 1
In Lesson 8.4, you learned that two triangles
are similar if two of their corresponding angles
are congruent. For example, ¤PQR ~ ¤STU.
Recall that the corresponding side lengths of
similar triangles are in proportion.
In the activity, you will see how a right triangle
can be divided into two similar right triangles.
Why you should learn it
RE
S
P
U
T
q
R
ACTIVITY
Developing
Concepts
Investigating Similar Right Triangles
1
Cut an index card along one of
its diagonals.
2
On one of the right triangles, draw
an altitude from the right angle to
the hypotenuse. Cut along the
altitude to form two right triangles.
3
You should now have three right
triangles. Compare the triangles.
What special property do they
share? Explain.
FE
You can use right
triangles and a geometric
mean to help you estimate
distances, such as finding the
approximate height of a
monorail track in
Example 3.
AL LI
PROPORTIONS IN RIGHT TRIANGLES
In the activity, you may have discovered the following theorem. A plan for
proving the theorem appears on page 528, and you are asked to prove it in
Exercise 34 on page 533.
THEOREM
THEOREM 9.1
C
If the altitude is drawn to the hypotenuse
of a right triangle, then the two triangles
formed are similar to the original triangle
and to each other.
A
D
B
¤CBD ~ ¤ABC, ¤ACD ~ ¤ABC,
and ¤CBD ~ ¤ACD
9.1 Similar Right Triangles
527
Page 2 of 8
A plan for proving Theorem 9.1 is shown below.
Æ
GIVEN ¤ABC is a right triangle; altitude CD is
C
Æ
drawn to hypotenuse AB.
PROVE ¤CBD ~ ¤ABC, ¤ACD ~ ¤ABC,
and ¤CBD ~ ¤ACD.
A
D
B
Plan for Proof First prove that ¤CBD ~ ¤ABC. Each triangle has a right angle,
and each includes ™B. The triangles are similar by the AA Similarity Postulate.
You can use similar reasoning to show that ¤ACD ~ ¤ABC. To show that
¤CBD ~ ¤ACD, begin by showing that ™ACD £ ™B because they are both
complementary to ™DCB. Then you can use the AA Similarity Postulate.
EXAMPLE 1
RE
FE
L
AL I
Finding the Height of a Roof
ROOF HEIGHT A roof has a cross section
Y
that is a right triangle. The diagram shows
the approximate dimensions of this cross section.
5.5 m
h
3.1 m
a. Identify the similar triangles.
Z
b. Find the height h of the roof.
6.3 m
W
SOLUTION
a. You may find it helpful to sketch the three similar right triangles so that the
corresponding angles and sides have the same orientation. Mark the
congruent angles. Notice that some sides appear in more than one triangle.
Æ
For instance, XY is the hypotenuse in ¤XYW and the shorter leg in ¤XZY.
Z
Z
Y
6.3 m
5.5 m
3.1 m
X
STUDENT HELP
Study Tip
In Example 1, all the side
lengths of ¤XZY are
given. This makes it a
good choice for setting
up a proportion to find an
unknown side length of
¤XYW.
528
5.5 m
h
W
Y
h
W
X
3.1 m
Y
¤XYW ~ ¤YZW ~ ¤XZY
b. Use the fact that ¤XYW ~ ¤XZY to write a proportion.
YW
XY
= ZY
XZ
Corresponding side lengths are in proportion.
h
3.1
= 5.5
6.3
Substitute.
6.3h = 5.5(3.1)
h ≈ 2.7
Cross product property
Solve for h.
The height of the roof is about 2.7 meters.
Chapter 9 Right Triangles and Trigonometry
X
Page 3 of 8
GOAL 2
USING A GEOMETRIC MEAN TO SOLVE PROBLEMS
Æ
In right ¤ABC, altitude CD is drawn to the
hypotenuse, forming two smaller right triangles
that are similar to ¤ABC. From Theorem 9.1,
you know that ¤CBD ~ ¤ACD ~ ¤ABC.
C
A
D
B
B
C
B
C
D
A
D
A
C
Æ
STUDENT HELP
Look Back
The geometric mean of
two numbers a and b is
the positive number x
a
x
x
b
such that = . For
more help with finding
a geometric mean,
see p. 466.
Notice that CD is the longer leg of ¤CBD and the shorter leg of ¤ACD. When
you write a proportion comparing the leg lengths of ¤CBD and ¤ACD, you can
see that CD is the geometric mean of BD and AD.
shorter leg of ¤CBD
shorter leg of ¤ACD
Æ
BD
CD
= CD
AD
longer leg of ¤CBD
longer leg of ¤ACD
Æ
Sides CB and AC also appear in more than one triangle. Their side lengths are
also geometric means, as shown by the proportions below:
hypotenuse of ¤ABC
hypotenuse of ¤CBD
hypotenuse of ¤ABC
hypotenuse of ¤ACD
AB
CB shorter leg of ¤ABC
=
CB
DB shorter leg of ¤CBD
AB
AC
= AC
AD
longer leg of ¤ABC
longer leg of ¤ACD
These results are expressed in the theorems below. You are asked to prove the
theorems in Exercises 35 and 36.
GEOMETRIC MEAN THEOREMS
GEOMETRIC MEAN THEOREMS
THEOREM 9.2
In a right triangle, the altitude from the
right angle to the hypotenuse divides the
hypotenuse into two segments.
The length of the altitude is the geometric
mean of the lengths of the two segments.
C
A
D
B
CD
BD
= AD
CD
THEOREM 9.3
In a right triangle, the altitude from the right
angle to the hypotenuse divides the hypotenuse
into two segments.
AB
CB
= CB
DB
The length of each leg of the right triangle is the
geometric mean of the lengths of the hypotenuse
and the segment of the hypotenuse that is adjacent
to the leg.
AC
AB
= AD
AC
9.1 Similar Right Triangles
529
Page 4 of 8
Using a Geometric Mean
EXAMPLE 2
STUDENT HELP
Skills Review
For help with
simplifying radicals,
see p. 799.
Study Tip
In part (a) of Example 2,
the equation 18 = x 2 has
two solutions, +18
and
º18
. Because you are
finding a length, you use
the positive square root.
Find the value of each variable.
a.
b.
2
y
x
6
5
3
SOLUTION
a. Apply Theorem 9.2.
b. Apply Theorem 9.3.
y
5+2
= 2
y
y
7
= 2
y
x
6
= 3
x
18 = x2
18 = x
14 = y2
9 • 2 = x
14 = y
32 = x
EXAMPLE 3
FOCUS ON
APPLICATIONS
Using Indirect Measurement
MONORAIL TRACK To estimate the height
Not drawn to scale
of a monorail track, your friend holds a
cardboard square at eye level. Your friend
lines up the top edge of the square with the
track and the bottom edge with the ground.
You measure the distance from the ground to
your friend’s eye and the distance from your
friend to the track.
RE
FE
L
AL I
MONORAILS
are railways that
have only one rail. The
Jacksonville monorail,
shown in the photo above,
can travel up to 35 miles
per hour and can carry
about 56 passengers
per car.
In the diagram, XY = h º 5.75 is the
difference between the track height h and
your friend’s eye level. Use Theorem 9.2
to write a proportion involving XY. Then
you can solve for h.
XY
WY
= WY
ZY
16
h º 5.75
= 5.75
16
5.75(h º 5.75) = 162
5.75h º 33.0625 = 256
5.75h = 289.0625
h ≈ 50
530
track
h – 5.75 ft
W
h
16 ft
Y
5.75 ft
Z
Geometric Mean Theorem 9.2
Substitute.
Cross product property
Distributive property
Add 33.0625 to each side.
Divide each side by 5.75.
The height of the track is about 50 feet.
Chapter 9 Right Triangles and Trigonometry
X
Page 5 of 8
GUIDED PRACTICE
Vocabulary Check
Concept Check
✓
✓
In Exercises 1–3, use the diagram at the right.
M
L
? of ML and JL.
1. In the diagram, KL is the 2. Complete the following statement:
? ~¤ ? .
¤JKL ~ ¤
K
J
3. Which segment’s length is the geometric mean
of ML and MJ?
Skill Check
✓
In Exercises 4–9, use the diagram above. Complete the proportion.
?
KM
4. = JK
KL
JM
JK
5. = ?
JL
?
LK
6. = LK
LM
JM
KM
7. = ?
LM
LK
JK
8. = LM
?
?
MK
9. = JK
MJ
10. Use the diagram at the right. Find DC.
F
Then find DF. Round decimals to the
nearest tenth.
72
65
D
97
E
C
PRACTICE AND APPLICATIONS
q
STUDENT HELP
SIMILAR TRIANGLES Use the diagram.
Extra Practice
to help you master
skills is on p. 819.
11. Sketch the three similar triangles in the
T
R
diagram. Label the vertices.
12. Write similarity statements for the
S
three triangles.
USING PROPORTIONS Complete and solve the proportion.
x
?
4
x
5
x
13. = 14. = 15. = 20
12
x
?
x
?
x
20
x
3
x
12
4
9
2
COMPLETING PROPORTIONS Write similarity statements for the three
similar triangles in the diagram. Then complete the proportion.
STUDENT HELP
QT
SQ
17. = SQ
?
?
XW
16. = YW
ZW
Y
HOMEWORK HELP
q
W
Example 1: Exs. 11–31
Example 2: Exs. 11–31
Example 3: Ex. 32
X
?
EG
18. = EG
EF
E
H
F
T
Z
S
R
G
9.1 Similar Right Triangles
531
Page 6 of 8
FINDING LENGTHS Write similarity statements for three triangles in the
diagram. Then find the given length. Round decimals to the nearest tenth.
19. Find DB.
20. Find HF.
21. Find JK.
G
C
M
20
12
A
16
D
E
B
22. Find QS.
q
H
32
F
25
J
23. Find CD.
40
15 K
L
24. Find FH.
C
R
F
25
32
7
G
T
S
A
4
D
4
B
E
H
xy USING ALGEBRA Find the value of each variable.
25.
26.
3
INT
STUDENT HELP
NE
ER T
28.
12
20
c
e
16
32
x
y
x9
8
24
18
d
31.
KITE DESIGN You are designing a diamond-
shaped kite. You know that AD = 44.8 centimeters,
DC = 72 centimeters, and AC = 84.8 centimeters.
Æ
You want to use a straight crossbar BD. About how
long should it be? Explain.
32.
532
m
30.
z
Visit our Web site
www.mcdougallittell.com
for help with Exs. 28–30.
7
29.
14
HOMEWORK HELP
5
x
9
x
27.
16
ROCK CLIMBING You and a friend
want to know how much rope you need
to climb a large rock. To estimate the
height of the rock, you use the method
from Example 3 on page 530. As shown
at the right, your friend uses a square to
line up the top and the bottom of the rock.
You measure the vertical distance from
the ground to your friend’s eye and the
distance from your friend to the rock.
Estimate the height of the rock.
Chapter 9 Right Triangles and Trigonometry
D
A
B
C
Not drawn to scale
h
18 ft
1
5 2 ft
Page 7 of 8
STUDENT HELP
Look Back
For help with finding
the area of a triangle,
see p. 51.
33. FINDING AREA Write similarity statements
B
for the three similar right triangles in the
diagram. Then find the area of each triangle.
Explain how you got your answers.
2.5 m D
1.5 m
2m
A
C
PROVING THEOREMS 9.1, 9.2, AND 9.3 In Exercises 34–36, use the
diagram at the right.
34. Use the diagram to prove Theorem 9.1 on page 527.
C
(Hint: Look back at the plan for proof on page 528.)
GIVEN ¤ABC is a right triangle;
Æ
Æ
altitude CD is drawn to hypotenuse AB.
A
D
B
PROVE ¤CBD ~ ¤ABC, ¤ACD ~ ¤ABC,
and ¤CBD ~ ¤ACD.
35. Use the diagram to prove Theorem 9.2 on page 529.
GIVEN ¤ABC is a right triangle;
Æ
Æ
altitude CD is drawn to hypotenuse AB.
BD
CD
CD
AD
PROVE = 36. Use the diagram to prove Theorem 9.3 on page 529.
Æ
GIVEN ¤ABC is a right triangle; altitude CD is
Æ
drawn to hypotenuse AB.
AB
BC
BC
BD
AB
AC
AC
AD
PROVE = and = USING TECHNOLOGY In Exercises 37–40, use geometry software.
You will demonstrate that Theorem 9.2 is true only for a right triangle.
Follow the steps below to construct a triangle.
1
Draw a triangle and label its vertices
A, B, and C. The triangle should not be
a right triangle.
Æ
C
Æ
2
Draw altitude CD from point C to side AB.
3
Measure ™C. Then measure AD, CD, and BD.
Æ Æ
Æ
A
D
B
BD
CD
37. Calculate the values of the ratios and .
CD
AD
What does Theorem 9.2 say about the values of
these ratios?
38. Drag point C until m™C = 90°. What happens to
BD
CD
the values of the ratios and ?
CD
AD
39. Explain how your answers to Exercises 37 and 38 support the conclusion that
Theorem 9.2 is true only for a right triangle.
40. Use the triangle you constructed to show that Theorem 9.3 is true only for a
right triangle. Describe your procedure.
9.1 Similar Right Triangles
533
Page 8 of 8
Test
Preparation
41. MULTIPLE CHOICE Use the diagram at the
C
right. Decide which proportions are true.
D
DB
DA
I. = DC
DB
BA
CB
II. = CB
BD
BA
CA
III. = CA
BA
DB
DA
IV. = BC
BA
A
¡
B
¡
I only
A
C
¡
II only
B
D
¡
I and II only
I and IV only
42. MULTIPLE CHOICE In the diagram above, AC = 24 and BC = 12.
★ Challenge
Find AD. If necessary, round to the nearest hundredth.
A 6
B 16.97
C 18
¡
¡
¡
43.
D
¡
20.78
Writing
Two methods for indirectly measuring the height of a building are
shown below. For each method, describe what distances need to be measured
directly. Explain how to find the height of the building using these
measurements. Describe one advantage and one disadvantage of each
method. Copy and label the diagrams as part of your explanations.
Method 1 Use the method
described in Example 3 on
page 530.
Method 2 Use the method
described in Exercises 55 and 56
on page 486.
B
R
Not drawn to scale
Not drawn to scale
N
A
EXTRA CHALLENGE
C
M P
D
q
S
www.mcdougallittell.com
MIXED REVIEW
xy SOLVING EQUATIONS Solve the equation. (Skills Review, p. 800, for 9.2)
44. n2 = 169
45. 14 + x 2 = 78
46. d 2 + 18 = 99
LOGICAL REASONING Write the converse of the statement. Decide
whether the converse is true or false. (Review 2.1)
47. If a triangle is obtuse, then one of its angles is greater than 90°.
48. If two triangles are congruent, then their corresponding angles are congruent.
FINDING AREA Find the area of the figure. (Review 1.7, 6.7 for 9.2)
49.
534
51.
50.
6 in.
4.5 cm
12 in.
7 cm
Chapter 9 Right Triangles and Trigonometry
12 m
5m
13 m
Page 1 of 7
9.2
The Pythagorean Theorem
What you should learn
GOAL 1 Prove the
Pythagorean Theorem.
GOAL 2 Use the
Pythagorean Theorem to
solve real-life problems,
such as determining how far
a ladder will reach in Ex. 32.
Why you should learn it
RE
FE
To measure real-life
lengths indirectly, such as
the length of the support
beam of a skywalk in
Example 4.
AL LI
GOAL 1
PROVING THE PYTHAGOREAN THEOREM
In this lesson, you will study one of the most famous theorems in mathematics—
the Pythagorean Theorem. The relationship it describes has been known for
thousands of years.
THEOREM
Pythagorean Theorem
THEOREM 9.4
c
a
In a right triangle, the square of the length of the
hypotenuse is equal to the sum of the squares of
the lengths of the legs.
b
2
c = a2 + b2
PROVING THE PYTHAGOREAN THEOREM There are many different proofs of
the Pythagorean Theorem. One is shown below. Other proofs are found in
Exercises 37 and 38 on page 540, and in the Math and History feature on page 557.
GIVEN In ¤ABC, ™BCA is a right angle.
PROVE a2 + b2 = c2
Æ
Plan for Proof Draw altitude CD to the
hypotenuse. Then apply Geometric Mean
Theorem 9.3, which states that when the
altitude is drawn to the hypotenuse of a
right triangle, each leg of the right triangle
is the geometric mean of the hypotenuse
and the segment of the hypotenuse that is
adjacent to that leg.
Statements
Proof
A
f
D
c
b
C
e
a
B
Reasons
1. Draw a perpendicular from
1. Perpendicular Postulate
Æ
C to AB.
c
c
a
b
2. = and = a
b
e
f
2. Geometric Mean Theorem 9.3
3. ce = a2 and cf = b2
3. Cross product property
4. ce + cf = a2 + b2
4. Addition property of equality
2
5. c(e + f ) = a + b
2
5. Distributive property
6. e + f = c
6. Segment Addition Postulate
7. c2 = a2 + b2
7. Substitution property of equality
9.2 The Pythagorean Theorem
535
Page 2 of 7
GOAL 2
USING THE PYTHAGOREAN THEOREM
A Pythagorean triple is a set of three positive integers a, b, and c that satisfy
the equation c2 = a2 + b2. For example, the integers 3, 4, and 5 form a
Pythagorean triple because 52 = 32 + 42.
Finding the Length of a Hypotenuse
EXAMPLE 1
Find the length of the hypotenuse of the right triangle.
Tell whether the side lengths form a Pythagorean triple.
12
5
x
SOLUTION
(hypotenuse)2 = (leg)2 + (leg)2
2
2
x = 5 + 12
2
Pythagorean Theorem
Substitute.
x2 = 25 + 144
Multiply.
x2 = 169
Add.
x = 13
Find the positive square root.
Because the side lengths 5, 12, and 13 are integers, they form a
Pythagorean triple.
..........
Many right triangles have side lengths that do not form a Pythagorean triple,
as shown in Example 2.
EXAMPLE 2
Finding the Length of a Leg
Find the length of the leg of the right triangle.
x
7
SOLUTION
(hypotenuse)2 = (leg)2 + (leg)2
STUDENT HELP
Skills Review
For help with simplifying
radicals, see p. 799.
14
Pythagorean Theorem
142 = 72 + x2
Substitute.
196 = 49 + x2
Multiply.
147 = x
2
147 = x
49 • 3 = x
73 = x
Subtract 49 from each side.
Find the positive square root.
Use product property.
Simplify the radical.
..........
In Example 2, the side length was written as a radical in simplest form. In reallife problems, it is often more convenient to use a calculator to write a decimal
approximation of the side length. For instance, in Example 2, x = 7 • 3 ≈ 12.1.
536
Chapter 9 Right Triangles and Trigonometry
Page 3 of 7
EXAMPLE 3
STUDENT HELP
Look Back
For help with finding
the area of a triangle,
see p. 51.
Finding the Area of a Triangle
Find the area of the triangle to the nearest tenth of a meter.
7m
SOLUTION
You are given that the base of the triangle is
10 meters, but you do not know the height h.
10 m
Because the triangle is isosceles, it can be divided into
two congruent right triangles with the given dimensions.
Use the Pythagorean Theorem to find the value of h.
72 = 52 + h2
Pythagorean Theorem
49 = 25 + h2
Multiply.
24 = h2
Subtract 25 from both sides.
24 = h
7m
h
7m
h
5m
Find the positive square root.
Now find the area of the original triangle.
1
2
1
= (10)(24)
2
Area = bh
24 m
≈ 24.5 m2
The area of the triangle is about 24.5 m2.
EXAMPLE 4
FOCUS ON
PEOPLE
10 m
Indirect Measurement
SUPPORT BEAM The skyscrapers shown on page 535 are connected by a
skywalk with support beams. You can use the Pythagorean Theorem to find
the approximate length of each support beam.
23.26 m 23.26 m
47.57 m
x
x
47.57 m
support
beams
RE
FE
L
AL I
CESAR PELLI
is an architect
who designed the twin
skyscrapers shown on
page 535. These 1483 foot
buildings tower over the city
of Kuala Lumpur, Malaysia.
Each support beam forms the hypotenuse of a right triangle. The right triangles
are congruent, so the support beams are the same length.
x2 = (23.26)2 + (47.57)2
Pythagorean Theorem
x = (2
3.2
6
)
+(47.5
7
)
Find the positive square root.
x ≈ 52.95
Use a calculator to approximate.
2
2
The length of each support beam is about 52.95 meters.
9.2 The Pythagorean Theorem
537
Page 4 of 7
GUIDED PRACTICE
✓
Concept Check ✓
Vocabulary Check
1. State the Pythagorean Theorem in your own words.
2. Which equations are true for ¤PQR?
A. r 2 = p2 + q2
2
2
2
2
2
2
B. q = p + r
C. p = r º q
q
r
p
R
D. r 2 = ( p + q)2
P
q
E. p2 = q2 + r 2
Skill Check
✓
Find the unknown side length. Tell whether the side lengths form
a Pythagorean triple.
3.
x
4.
x
5.
x
4
1
8
10
8
2
6.
ANEMOMETER An anemometer (an uh MAHM ih tur)
is a device used to measure windspeed. The anemometer
shown is attached to the top of a pole. Support wires are
attached to the pole 5 feet above the ground. Each support
wire is 6 feet long. How far from the base of the pole is
each wire attached to the ground?
6 ft
5 ft
d
PRACTICE AND APPLICATIONS
STUDENT HELP
Extra Practice
to help you master
skills is on p. 819.
FINDING SIDE LENGTHS Find the unknown side length. Simplify answers
that are radicals. Tell whether the side lengths form a Pythagorean triple.
7.
8.
9
x
65
9.
x
6
39
89
x
72
10.
11.
x
12.
7
x
x
3
40
9
9
2
STUDENT HELP
HOMEWORK HELP
Example 1:
Example 2:
Example 3:
Example 4:
538
Exs. 7–24
Exs. 7–24
Exs. 25–30
Exs. 31–36
13.
x
8
14.
16
Chapter 9 Right Triangles and Trigonometry
15.
29
x
x
20
14
Page 5 of 7
FINDING LENGTHS Find the value of x. Simplify answers that are radicals.
16.
17.
18.
x
x
10
6
5
x
3
11
8
12
8
PYTHAGOREAN TRIPLES The variables r and s represent the lengths of the
legs of a right triangle, and t represents the length of the hypotenuse. The
values of r, s, and t form a Pythagorean triple. Find the unknown value.
STUDENT HELP
Look Back
For help with finding
areas of quadrilaterals,
see pp. 372–375.
19. r = 12, s = 16
20. r = 9, s = 12
21. r = 18, t = 30
22. s = 20, t = 101
23. r = 35, t = 37
24. t = 757, s = 595
FINDING AREA Find the area of the figure. Round decimal answers to the
nearest tenth.
25.
26.
27.
14 m
12 cm
8 cm
5m
7 cm
9 cm
28.
29.
5m
8.5 m
4m
8 cm
30.
10 cm
13 m
10 cm
12 m
8m
16 cm
31.
32.
SOFTBALL DIAMOND In slowpitch softball, the distance between
consecutive bases is 65 feet. The
pitcher’s plate is located on a line
between second base and home plate,
50 feet from home plate. How far is
the pitcher’s plate from second base?
Justify your answer.
second
65 ft
pitcher’s
plate
first
50 ft
third
65 ft
home
SAFETY The distance of the base of a ladder from the wall it leans
against should be at least 14 of the ladder’s total length. Suppose a 10 foot
ladder is placed according to these guidelines. Give the minimum distance of
the base of the ladder from the wall. How far up the wall will the ladder
reach? Explain. Include a sketch with your explanation.
33.
ART GALLERY You want to hang
a painting 3 feet from a hook near the
ceiling of an art gallery, as shown. In
addition to the length of wire needed for
hanging, you need 16 inches of wire to
secure the wire to the back of the painting.
Find the total length of wire needed to
hang the painting.
Need 8 in.
to secure
3 ft
Need 8 in.
to secure
2 ft 6 in.
9.2 The Pythagorean Theorem
539
Page 6 of 7
FOCUS ON
CAREERS
FE
RE
TRANS-ALASKA PIPELINE Metal
expands and contracts with changes in
temperature. The Trans-Alaska pipeline
was built to accommodate expansion and
contraction. Suppose that it had not been
built this way. Consider a 600 foot section
of pipe that expands 2 inches and buckles,
as shown below. Estimate the height h
of the buckle.
300 f t 1 in.
L
AL I
300 f t 1 in.
MECHANICAL
ENGINEERS
use science, mathematics,
and engineering principles
in their work. They evaluate,
install, operate, and maintain
mechanical products and
systems, such as the TransAlaska pipeline.
INT
34.
h
300 f t
Not drawn
to scale
300 f t
WRAPPING A BOX In Exercises 35 and 36, two methods are used to
wrap ribbon around a rectangular box with the dimensions shown below.
The amount of ribbon needed does not include a knot or bow.
NE
ER T
Method 1
CAREER LINK
12 in.
Method 2
3 in.
www.mcdougallittell.com
6 in.
3 in.
12 in.
6 in.
35. How much ribbon is needed to
wrap the box using Method 1?
top
36. The red line on the diagram
at the right shows the path
the ribbon follows around
the box when Method 2 is
used. Does Method 2 use
more or less ribbon than
Method 1? Explain your
thinking.
INT
STUDENT HELP
NE
ER T
37.
back
bottom
top
front
left
side
Visit our Web site
www.mcdougallittell.com
for help with the proofs in
Exs. 37 and 38.
38.
bottom
PROVING THE PYTHAGOREAN THEOREM
Explain how the diagram at the right can be used
to prove the Pythagorean Theorem algebraically.
(Hint: Write two different expressions that represent
the area of the large square. Then set them equal to
each other.)
HOMEWORK HELP
right
side
a
b
b
c
a
a
c
c
b
c
b
a
GARFIELD’S PROOF James Abram Garfield, the twentieth president of
the United States, discovered a proof of the Pythagorean Theorem in 1876.
His proof involved the fact that a trapezoid can be formed from two
congruent right triangles and an isosceles right triangle.
Use the diagram to write a paragraph proof
showing that a2 + b2 = c2. (Hint: Write two
different expressions that represent the area of
the trapezoid. Then set them equal to each other.)
540
Chapter 9 Right Triangles and Trigonometry
b
c
c
a
b
a
Page 7 of 7
Test
Preparation
39. MULTI-STEP PROBLEM To find the length of a
D
diagonal of a rectangular box, you can use the
Pythagorean Theorem twice. Use the theorem
once with right ¤ABC to find the length of the
diagonal of the base.
A
C
)
+(B
C
)
AB = (A
2
2
Then use the theorem with right ¤ABD to
find the length of the diagonal of the box.
B
C
BD = (A
B
)2
+(A
D
)2
a. Is it possible to carry a 9 foot piece of lumber in an enclosed rectangular
trailer that is 4 feet by 8 feet by 4 feet?
b. Is it possible to store a 20 foot long pipe in a rectangular room that is
10 feet by 12 feet by 8 feet? Explain.
c.
★ Challenge
Writing
Write a formula for finding the diagonal d of a rectangular box
with length ¬, width w, and height h. Explain your reasoning.
PERIMETER OF A RHOMBUS The diagonals of a rhombus have lengths
a and b. Use this information in Exercises 40 and 41.
40. Prove that the perimeter of the rhombus is 2a
2+
b2.
EXTRA CHALLENGE
www.mcdougallittell.com
41. The perimeter of a rhombus is 80 centimeters. The lengths of its diagonals
are in the ratio 3 :4. Find the length of each diagonal.
MIXED REVIEW
USING RADICALS Evaluate the expression. (Algebra Review, p. 522, for 9.3)
42. (6
)2
43. (9
)2
44. (1
4)2
45. (22
)2
46. (41
3)2
47. º(54
9)2
48. 4(9
)2
49. (º73
)2
LOGICAL REASONING Determine whether the true statement can be
combined with its converse to form a true biconditional statement.
(Review 2.2)
50. If a quadrilateral is a square, then it has four congruent sides.
51. If a quadrilateral is a kite, then it has two pairs of congruent sides.
52. For all real numbers x, if x ≥ 1, then x2 ≥ 1.
1
53. For all real numbers x, if x > 1, then < 1.
x
54. If one interior angle of a triangle is obtuse, then the sum of the other two
interior angles is less than 90°.
xy USING ALGEBRA Prove that the points represent the vertices of a
parallelogram. (Review 6.3)
55. P(4, 3), Q(6, º8), R(10, º3), S(8, 8)
56. P(5, 0), Q(2, 9), R(º6, 6), S(º3, º3)
9.2 The Pythagorean Theorem
541
Page 1 of 1
ACTIVITY 9.3
Using Technology
Geometry Software Activity for use with Lesson 9.3
Investigating
Sides and Angles of Triangles
You can use geometry software to explore how the angle measures of a
triangle are related to its side lengths.
INT
STUDENT HELP
NE
ER T
CONSTRUCT
1 Construct a triangle. Label
SOFTWARE
HELP
Visit our Web site
www.mcdougallittell.com
to see instructions for
several different
software applications.
Æ Æ
C
98.07
2.11 cm
the vertices A, B, and C.
A
Æ
2 Measure AC, BC, and AB.
2.59 cm
3.56 cm
3 Calculate the value of (AC)2 + (BC)2.
Calculate the value of (AB)2.
(AC ) 2 (BC ) 2 11.16
(AB) 2 12.67
4 Measure ™C.
B
INVESTIGATE
1. Make a table like the one shown. Record the side lengths, the value of
(AC)2 + (BC)2, the value of (AB)2, and the measure of ™C. Round
decimals to the nearest hundredth.
AC
BC
AB
(AC)2 + (BC)2
(AB)2
m™C
Triangle 1
2.11 cm
2.59 cm
3.56 cm
11.16 cm
12.67 cm
98.07°
Triangle 2
?
?
?
?
?
?
Triangle 3
?
?
?
?
?
?
Triangle 4
?
?
?
?
?
?
Triangle 5
?
?
?
?
?
?
Triangle 6
?
?
?
?
?
?
2. Drag point C to change the measure of ™C and the shape of ¤ABC.
Find and record the values in the table.
3. Repeat Step 2 for several more triangles. In your table, be sure to
include some triangles for which ™C is an acute angle, some for
which ™C is a right angle, and some for which ™C is an obtuse angle.
CONJECTURE
4. Use the data in the table to look for a pattern. Make a conjecture
about how the value of (AC)2 + (BC)2 compares with the value
of (AB)2 when the measure of ™C is less than 90°, equal to 90°,
and greater than 90°.
542
Chapter 9 Right Triangles and Trigonometry
Page 1 of 7
9.3
The Converse of
the Pythagorean Theorem
What you should learn
GOAL 1 Use the Converse
of the Pythagorean Theorem
to solve problems.
GOAL 2 Use side lengths to
classify triangles by their
angle measures.
Why you should learn it
RE
FE
To determine whether
real-life angles are right
angles, such as the four
angles formed by the
foundation of a building
in Example 3.
AL LI
GOAL 1
USING THE CONVERSE
In Lesson 9.2, you learned that if a triangle is a right triangle, then the square of
the length of the hypotenuse is equal to the sum of the squares of the lengths of
the legs. The Converse of the Pythagorean Theorem is also true, as stated below.
Exercise 43 asks you to prove the Converse of the Pythagorean Theorem.
THEOREM
Converse of the Pythagorean Theorem
THEOREM 9.5
If the square of the length of the longest
side of a triangle is equal to the sum of the
squares of the lengths of the other two
sides, then the triangle is a right triangle.
2
2
B
c
a
C
A
b
2
If c = a + b , then ¤ABC is a right triangle.
You can use the Converse of the Pythagorean Theorem to verify that a given
triangle is a right triangle, as shown in Example 1.
EXAMPLE 1
Verifying Right Triangles
The triangles below appear to be right triangles. Tell whether they are right
triangles.
a.
b.
7
8
495
113
15
36
SOLUTION
Let c represent the length of the longest side of the triangle. Check to see whether
the side lengths satisfy the equation c2 = a2 + b2.
a. (1
13 )2 · 72 + 82
113 · 49 + 64
113 = 113 ✓
b.
(495)2 · 152 + 362
42 • (95 )2 · 152 + 362
16 • 95 · 225 + 1296
The triangle is a right triangle.
1520 ≠ 1521
The triangle is not a right triangle.
9.3 The Converse of the Pythagorean Theorem
543
Page 2 of 7
GOAL 2
CLASSIFYING TRIANGLES
Sometimes it is hard to tell from looking whether a triangle is obtuse or acute.
The theorems below can help you tell.
THEOREMS
THEOREM 9.6
A
If the square of the length of the longest side of a
triangle is less than the sum of the squares of the lengths
of the other two sides, then the triangle is acute.
c
b
a
C
If c 2 < a 2 + b 2, then ¤ABC is acute.
B
c2 < a2 + b2
THEOREM 9.7
If the square of the length of the longest side of a
triangle is greater than the sum of the squares of
the lengths of the other two sides, then the triangle
is obtuse.
2
2
A
2
If c > a + b , then ¤ABC is obtuse.
THEOREMS
EXAMPLE 2
c
b
C
B
a
2
2
c >a +b
2
Classifying Triangles
Decide whether the set of numbers can represent the side lengths of a
triangle. If they can, classify the triangle as right, acute, or obtuse.
a. 38, 77, 86
b. 10.5, 36.5, 37.5
SOLUTION
STUDENT HELP
Look Back
For help with the
Triangle Inequality,
see p. 297.
You can use the Triangle Inequality to confirm that each set of numbers can
represent the side lengths of a triangle.
Compare the square of the length of the longest side with the sum of the squares
of the lengths of the two shorter sides.
? a2 + b2
c2 a.
? 382 + 772
862 Multiply.
7396 > 7373
c 2 is greater than a 2 + b 2.
Because c2 > a2 + b2, the triangle is obtuse.
? 10.52 + 36.52
37.52 Compare c 2 with a 2 + b 2.
Substitute.
? 110.25 + 1332.25
1406.25 Multiply.
1406.25 < 1442.5
c 2 is less than a 2 + b 2.
544
Substitute.
? 1444 + 5929
7396 ? a2 + b2
c2 b.
Compare c 2 with a 2 + b 2.
Because c2 < a2 + b2, the triangle is acute.
Chapter 9 Right Triangles and Trigonometry
Page 3 of 7
EXAMPLE 3
RE
FE
L
AL I
Building a Foundation
CONSTRUCTION You use four stakes and string to mark the foundation
of a house. You want to make sure the foundation is rectangular.
a. A friend measures the four sides to be 30 feet,
30 feet, 72 feet, and 72 feet. He says these
measurements prove the foundation is
rectangular. Is he correct?
30 ft
72 ft
78 ft
72 ft
b. You measure one of the diagonals to
be 78 feet. Explain how you can use
this measurement to tell whether the
foundation will be rectangular.
30 ft
SOLUTION
a. Your friend is not correct. The foundation
STUDENT HELP
Look Back
For help with classifying
quadrilaterals, see
Chapter 6.
could be a nonrectangular parallelogram,
as shown at the right.
72 ft
30 ft
72 ft
b. The diagonal divides the foundation into
two triangles. Compare the square of the
length of the longest side with the sum of
the squares of the shorter sides of one of
these triangles. Because 302 + 722 = 782,
you can conclude that both the triangles
are right triangles.
30 ft
72 ft
30 ft
78 ft
30 ft
72 ft
The foundation is a parallelogram with two right angles, which implies
that it is rectangular.
GUIDED PRACTICE
Vocabulary Check
✓
1. State the Converse of the Pythagorean Theorem in
your own words.
Concept Check
✓
c
2. Use the triangle shown at the right. Find values for
c so that the triangle is acute, right, and obtuse.
Skill Check
✓
18
24
In Exercises 3–6, match the side lengths with the appropriate description.
3. 2, 10, 11
A. right triangle
4. 13, 5, 7
B. acute triangle
5. 5, 11, 6
C. obtuse triangle
6. 6, 8, 10
D. not a triangle
7.
22 cm
KITE DESIGN You are making the
diamond-shaped kite shown at the right. You
measure the crossbars to determine whether
they are perpendicular. Are they? Explain.
38 cm
45 cm
9.3 The Converse of the Pythagorean Theorem
545
Page 4 of 7
PRACTICE AND APPLICATIONS
STUDENT HELP
Extra Practice
to help you master
skills is on pp. 819
and 820.
VERIFYING RIGHT TRIANGLES Tell whether the triangle is a right triangle.
8.
9.
97
65
10.
20.8
89
39
23
80
72
11.
12.
26
1
10.5
13.
5
2
5
20
33
13
435
CLASSIFYING TRIANGLES Decide whether the numbers can represent the
side lengths of a triangle. If they can, classify the triangle as right, acute,
or obtuse.
14. 20, 99, 101
15. 21, 28, 35
16. 26, 10, 17
17. 2, 10, 12
18. 4, 6
7, 9
19. 1
3, 6, 7
20. 16, 30, 34
21. 10, 11, 14
22. 4, 5, 5
23. 17, 144, 145
24. 10, 49, 50
25. 5
, 5, 5.5
CLASSIFYING QUADRILATERALS Classify the quadrilateral. Explain how
you can prove that the quadrilateral is that type.
26.
27.
14
265
28.
5
3
4
8
1
2
CHOOSING A METHOD In Exercises 29–31, you will use two different
methods for determining whether ¤ABC is a right triangle.
29.
30.
Æ
Find the slope of AC and the slope
of BC. What do the slopes tell you about ™ACB?
Is ¤ABC a right triangle? How do you know?
Method 1
HOMEWORK HELP
Example 1: Exs. 8–13, 30
Example 2: Exs. 14–28,
31–35
Example 3: Exs. 39, 40
whether a given triangle is right, acute, or
obtuse? Explain.
A(4, 6)
C(0, 3)
1
1
xy USING ALGEBRA Graph points P, Q, and R. Connect the points to form
¤PQR. Decide whether ¤PQR is right, acute, or obtuse.
32. P(º3, 4), Q(5, 0), R(º6, º2)
546
B(3, 7)
Method 2 Use the Distance Formula and
the Converse of the Pythagorean Theorem
to determine whether ¤ABC is a right triangle.
31. Which method would you use to determine
STUDENT HELP
y
Æ
Chapter 9 Right Triangles and Trigonometry
33. P(º1, 2), Q(4, 1), R(0, º1)
x
Page 5 of 7
PROOF Write a proof.
34. GIVEN AB = 3, BC = 2,
35. GIVEN AB = 4, BC = 2,
AC = 10
AC = 4
PROVE ™1 is acute.
PROVE ™1 is acute.
B 1
A
B 1
4
3
2
2
10
C
INT
STUDENT HELP
NE
ER T
36.
HOMEWORK HELP
Visit our Web site
www.mcdougallittell.com
for help with Ex. 36.
4
A
C
PROOF Prove that if a, b, and c are a Pythagorean triple, then ka, kb, and
kc (where k > 0) represent the side lengths of a right triangle.
37. PYTHAGOREAN TRIPLES Use the results of Exercise 36 and the
Pythagorean triple 5, 12, 13. Which sets of numbers can represent the side
lengths of a right triangle?
1
1
A. 50, 120, 130
B. 20, 48, 56
C. 1, 3, 3 D. 1, 2.4, 2.6
4
38.
4
TECHNOLOGY Use geometry software to
construct each of the following figures: a
nonspecial quadrilateral, a parallelogram, a
rhombus, a square, and a rectangle. Label the
sides of each figure a, b, c, and d. Measure each
side. Then draw the diagonals of each figure and
label them e and f. Measure each diagonal. For
which figures does the following statement
appear to be true?
c
b
f
e
d
a
a2 + b2 + c 2 + d 2 = e2 + f 2
FOCUS ON
APPLICATIONS
RE
FE
L
AL I
BABYLONIAN
TABLET
This photograph shows part
of a Babylonian clay tablet
made around 350 B.C. The
tablet contains a table of
numbers written in
cuneiform characters.
39.
40.
HISTORY CONNECTION The Babylonian
tablet shown at the left contains several sets
of triangle side lengths, suggesting that the
Babylonians may have been aware of the
relationships among the side lengths of right,
triangles. The side lengths in the table at the
right show several sets of numbers from the
tablet. Verify that each set of side lengths
forms a Pythagorean triple.
AIR TRAVEL You take off in a jet from
Cincinnati, Ohio, and fly 403 miles due east
to Washington, D.C. You then fly 714 miles
to Tallahassee, Florida. Finally, you fly
599 miles back to Cincinnati. Is Cincinnati
directly north of Tallahassee? If not, how
would you describe its location relative
to Tallahassee?
a
b
c
120
119
169
4,800
4,601
6,649
13,500
12,709
18,541
Cincinnati 403 mi Washington,
D.C.
599 mi
714 mi
N
W
Tallahassee
9.3 The Converse of the Pythagorean Theorem
E
S
547
Page 6 of 7
A
DEVELOPING PROOF Complete the
41.
proof of Theorem 9.6 on page 544.
c
GIVEN In ¤ABC, c2 < a 2 + b 2.
PROVE ¤ABC is an acute triangle.
B
b
a
œ
C
x
b
Plan for Proof Draw right ¤PQR with side
lengths a, b, and x. Compare lengths c and x.
Statements
2
2
2
2
2
2. c < a + b
R
? 1. ? 2. ? 3. 4. A property of square roots
3. c 2 < x 2
4. c < x
6. ™C is an acute angle.
? 5. ? 6. 7. ¤ABC is an acute triangle.
? 7. 5. m™C < m™R
42.
a
Reasons
2
1. x = a + b
P
PROOF Prove Theorem 9.7 on page 544. Include a diagram and Given
and Prove statements. (Hint: Look back at Exercise 41.)
PROOF Prove the Converse of the Pythagorean Theorem.
43.
Æ
GIVEN In ¤LNM, LM is the longest side;
M
c 2 = a 2 + b 2.
c
PROVE ¤LNM is a right triangle.
Plan for Proof Draw right ¤PQR with
a
L
N
side lengths a, b, and x. Compare lengths
c and x.
Test
Preparation
P
A
¡
B
¡
C
¡
D
¡
The quantity in column A is greater.
The quantity in column B is greater.
The two quantities are equal.
The relationship cannot be determined
from the given information.
548
a
R
E
B
77
91
82
85
A
36
C
44.
m™A
m™D
45.
m™B + m™C
m™E + m™F
46.
PROOF Prove the converse of Theorem 9.2 on page 529.
D
Chapter 9 Right Triangles and Trigonometry
F
N
t
t is the geometric mean of r and s.
PROVE ¤MQN is a right triangle.
40
Æ
GIVEN In ¤MQN, altitude NP is drawn to MQ ;
www.mcdougallittell.com
b
Column B
Æ
EXTRA CHALLENGE
x
QUANTITATIVE COMPARISON Choose the statement that is true about the
given quantities.
Column A
★ Challenge
œ
b
M
s
P r
q
Page 7 of 7
MIXED REVIEW
SIMPLIFYING RADICALS Simplify the expression. (Skills Review, p. 799, for 9.4)
47. 2
2 • 2
48. 6
• 8
49. 1
4 • 6
50. 1
5 • 6
3
51. 1
1
4
52. 5
12
53. 1
8
8
54. 2
4
DILATIONS Identify the dilation and find its scale factor. (Review 8.7)
55.
56.
16
40
L
F’
28
F
24
L’
M’
C
H
G’
G
C
H’
N’
57. xy USING ALGEBRA In the diagram,
Æ
˘
M
N
P
Æ
PS bisects ™RPT, and PS is the perpendicular
Æ
bisector of RT. Find the values of x and y.
x 36
5x
2y
(Review 5.1)
R
QUIZ 1
y 11 T
S
Self-Test for Lessons 9.1–9.3
In Exercises 1–4, use the diagram. (Lesson 9.1)
1. Write a similarity statement about the three
C
triangles in the diagram.
9
D
2. Which segment’s length is the geometric mean
15
of CD and AD?
3. Find AC.
A
4. Find BD.
B
Find the unknown side length. Simplify answers that are radicals.
(Lesson 9.2)
5.
6.
7
3
7.
x
x
12
6
18
x
CITY PARK The diagram shown at
the right shows the dimensions of a
triangular city park. Does this city park
have a right angle? Explain. (Lesson 9.3)
d
9y
21
Park
140 yd
8.
18
168 yd
9.3 The Converse of the Pythagorean Theorem
549
Page 1 of 1
ACTIVITY 9.4
Developing Concepts
GROUP ACTIVITY
Work in a group of three
students.
MATERIALS
• paper
• pencil
• ruler
• compass
Investigating
Special Right Triangles
Group Activity for use with Lesson 9.4
QUESTION A triangle with angle measures of 45°, 45°, and 90°, or 30°,
60°, and 90° is called a special right triangle. How are the side lengths of a
45°-45°-90° triangle or a 30°-60°-90° triangle related?
EXPLORING THE CONCEPT: 45°-45°-90° TRIANGLE
1. Construct an isosceles right triangle. The
length of a leg of the triangle should be 3,
4, or 5 centimeters. Each person in your
group should choose a different length.
P
2. Use the Pythagorean Theorem to find the
length of the hypotenuse. Write the length
in simplest radical form.
q
CONJECTURE
R
3. Compare your results with those of the other
students in your group. What pattern do you
observe? Make a conjecture about the
relationships among the side lengths of an
isosceles right triangle.
EXPLORING THE CONCEPT: 30°-60°-90° TRIANGLE
4. Construct an equilateral triangle with side
C
lengths of 4, 6, or 8 centimeters. Each person
in your group should choose a different length.
5. Next construct the altitude from one of the
vertices. The equilateral triangle is now
divided into two congruent right triangles
whose angle measures are 30°-60°-90°.
A
D
B
6. Find the side lengths of one of the right
triangles. Write each length in simplest
radical form.
CONJECTURE
7. Find each of the following ratios:
•
•
the length of the hypotenuse to the length of the shorter leg
the length of the longer leg to the length of the shorter leg
Compare your ratios with those of the other students in your group. What
patterns do you observe? Make a conjecture about the ratios of the side
lengths of a 30°-60°-90° triangle.
550
Chapter 9 Right Triangles and Trigonometry
Page 1 of 7
9.4
Special Right Triangles
What you should learn
GOAL 1 Find the side
lengths of special right
triangles.
GOAL 2 Use special right
triangles to solve real-life
problems, such as finding the
side lengths of the triangles
in the spiral quilt design
in Exs. 31–34.
Why you should learn it
RE
SIDE LENGTHS OF SPECIAL RIGHT TRIANGLES
Right triangles whose angle measures are 45°-45°-90° or 30°-60°-90° are called
special right triangles. In the Activity on page 550, you may have noticed
certain relationships among the side lengths of each of these special right
triangles. The theorems below describe these relationships. Exercises 35 and 36
ask you to prove the theorems.
THEOREMS ABOUT SPECIAL RIGHT TRIANGLES
THEOREM 9.8
45°-45°-90° Triangle Theorem
In a 45°-45°-90° triangle, the hypotenuse
is 2
times as long as each leg.
2x
45
x
45
x
Hypotenuse = 2
• leg
FE
To use special right
triangles to solve real-life
problems, such as finding the
height of a tipping platform
in Example 4.
AL LI
GOAL 1
THEOREM 9.9
30°-60°-90° Triangle Theorem
In a 30°-60°-90° triangle, the hypotenuse
is twice as long as the shorter leg, and
the longer leg is 3
times as long as the
shorter leg.
60
2x
x
30
3 x
Hypotenuse = 2 • shorter leg
Longer leg = 3
• shorter leg
EXAMPLE 1
Finding the Hypotenuse in a 45°-45°-90° Triangle
Find the value of x.
3
SOLUTION
By the Triangle Sum Theorem, the measure
of the third angle is 45°. The triangle is a
45°-45°-90° right triangle, so the length x
of the hypotenuse is 2 times the length
of a leg.
Hypotenuse = 2 • leg
3
45
x
45°-45°-90° Triangle Theorem
x = 2 • 3
Substitute.
x = 32
Simplify.
9.4 Special Right Triangles
551
Page 2 of 7
EXAMPLE 2
Finding a Leg in a 45°-45°-90° Triangle
5
Find the value of x.
SOLUTION
x
x
Because the triangle is an isosceles right triangle,
its base angles are congruent. The triangle is a
45°-45°-90° right triangle, so the length of the
hypotenuse is 2 times the length x of a leg.
Hypotenuse = 2 • leg
5 = 2 • x
5
x
2
=
2
2
5
=x
2
2
• 5 =x
2
2
52
=x
2
EXAMPLE 3
INT
STUDENT HELP
NE
ER T
HOMEWORK HELP
Visit our Web site
www.mcdougallittell.com
for extra examples.
45°-45°-90° Triangle Theorem
Substitute.
Divide each side by 2.
Simplify.
Multiply numerator and denominator by 2.
Simplify.
Side Lengths in a 30°-60°-90° Triangle
Find the values of s and t.
60
t
SOLUTION
Because the triangle is a 30°-60°-90° triangle,
the longer leg is 3 times the length s of the
shorter leg.
Longer leg = 3 • shorter leg
5 = 3 • s
3
•s
5
=
3
3
5
=s
3
3
5
•=s
3
3
53
=s
3
30
5
30°-60°-90° Triangle Theorem
Substitute.
Divide each side by 3.
Simplify.
Multiply numerator and denominator by 3.
Simplify.
The length t of the hypotenuse is twice the length s of the shorter leg.
Hypotenuse = 2 • shorter leg
53
3
552
30°-60°-90° Triangle Theorem
t=2•
Substitute.
103
t=
Simplify.
3
Chapter 9 Right Triangles and Trigonometry
s
Page 3 of 7
GOAL 2
USING SPECIAL RIGHT TRIANGLES IN REAL LIFE
EXAMPLE 4
RE
FE
L
AL I
Finding the Height of a Ramp
TIPPING PLATFORM A tipping platform is a ramp used to unload trucks,
as shown on page 551. How high is the end of an 80 foot ramp when it is
tipped by a 30° angle? by a 45° angle?
ramp
height
of ramp
80 f t
angle of
elevation
SOLUTION
When the angle of elevation is 30°, the height h of the ramp is the length of the
shorter leg of a 30°-60°-90° triangle. The length of the hypotenuse is 80 feet.
80 = 2h
30°-60°-90° Triangle Theorem
40 = h
Divide each side by 2.
When the angle of elevation is 45°, the height of the ramp is the length of a leg of
a 45°-45°-90° triangle. The length of the hypotenuse is 80 feet.
80 = 2 • h
45°-45°-90° Triangle Theorem
80
=h
2
Divide each side by 2.
56.6 ≈ h
Use a calculator to approximate.
When the angle of elevation is 30°, the ramp height is 40 feet. When the angle
of elevation is 45°, the ramp height is about 56 feet 7 inches.
EXAMPLE 5
RE
FE
L
AL I
Finding the Area of a Sign
ROAD SIGN The road sign is shaped like an
equilateral triangle. Estimate the area of the sign
by finding the area of the equilateral triangle.
SOLUTION
First find the height h of the triangle by dividing it into two
30°-60°-90° triangles. The length of the longer leg of one of
these triangles is h. The length of the shorter leg is 18 inches.
h = 3 • 18 = 183
18 in.
h
30°-60°-90° Triangle Theorem
36 in.
Use h = 183 to find the area of the equilateral triangle.
1
2
1
2
Area = bh = (36) 183 ≈ 561.18
The area of the sign is about 561 square inches.
9.4 Special Right Triangles
553
Page 4 of 7
GUIDED PRACTICE
✓
Concept Check ✓
Vocabulary Check
1. What is meant by the term special right triangles?
2. CRITICAL THINKING Explain why any two 30°-60°-90° triangles are similar.
Use the diagram to tell whether the equation is true or false.
Skill Check
✓
3. t = 73
4. t = 3
h
5. h = 2t
6. h = 14
h
7. 7 = 2
t
8. 7 = 3
60
h
7
30
t
Find the value of each variable. Write answers in simplest
radical form.
9.
10.
x
45
11.
a
60
4
45
9
k
4
b
h
PRACTICE AND APPLICATIONS
STUDENT HELP
Extra Practice
to help you master
skills is on p. 820.
xy USING ALGEBRA Find the value of each variable.
Write answers in simplest radical form.
12.
13.
x
14.
60
y
b
2
2
12
45
e
a
5
15.
16.
17.
10
d
8
30
45
20.
6
h
m
60
6
r
19.
p
f
n
62
45
30
p
16
45
d
c
18.
q
c
8
n
STUDENT HELP
HOMEWORK HELP
Example 1:
Example 2:
Example 3:
Example 4:
Exs. 12–23
Exs. 12–23
Exs. 12–23
Exs. 28–29,
34
Example 5: Exs. 24–27
FINDING LENGTHS Sketch the figure that is described. Find the requested
length. Round decimals to the nearest tenth.
21. The side length of an equilateral triangle is 5 centimeters. Find the length of
an altitude of the triangle.
22. The perimeter of a square is 36 inches. Find the length of a diagonal.
23. The diagonal of a square is 26 inches. Find the length of a side.
554
Chapter 9 Right Triangles and Trigonometry
Page 5 of 7
FINDING AREA Find the area of the figure. Round decimal answers to the
nearest tenth.
24.
25.
8 ft
26.
60
5m
12 ft
4m
4m
60
5m
27.
AREA OF A WINDOW A hexagonal window
consists of six congruent panes of glass. Each
pane is an equilateral triangle. Find the area of
the entire window.
8 ft
JEWELRY Estimate the length x of each earring.
28.
29.
30.
TOOLS Find the values of x and y for
s
the hexagonal nut shown at the right when
s = 2 centimeters. (Hint: In Exercise 27
above, you saw that a regular hexagon can
be divided into six equilateral triangles.)
y
x
LOGICAL REASONING The quilt design in the photo is based on the
pattern in the diagram below. Use the diagram in Exercises 31–34.
1
1
1
1
s
1
t
u
r
v
1
w
1
Wheel of Theodorus
31. Find the values of r, s, t, u, v, and w. Explain the procedure you used to
find the values.
32. Which of the triangles, if any, is a 45°-45°-90° triangle?
33. Which of the triangles, if any, is a 30°-60°-90° triangle?
34. xy USING ALGEBRA Suppose there are n triangles in the spiral. Write an
expression for the hypotenuse of the nth triangle.
9.4 Special Right Triangles
555
Page 6 of 7
35.
PARAGRAPH PROOF Write a paragraph
proof of Theorem 9.8 on page 551.
D
45
GIVEN ¤DEF is a 45°-45°-90° triangle.
PROVE The hypotenuse is 2
times as long
as each leg.
36.
E
PARAGRAPH PROOF Write a paragraph
proof of Theorem 9.9 on page 551.
GIVEN ¤ABC is a 30°-60°-90° triangle.
B
PROVE The hypotenuse is twice as long as
a
the shorter leg and the longer leg
is 3 times as long as the shorter leg.
Test
Preparation
45
F
60
30
C
Plan for Proof Construct ¤ADC congruent to
a
¤ABC. Then prove that ¤ABD is equilateral.
Express the lengths AB and AC in terms of a.
D
A
37. MULTIPLE CHOICE Which of the statements
below is true about the diagram at the right?
A
¡
C
¡
E
¡
B
¡
D
¡
x < 45
x > 45
x = 45
21
x ≤ 45
x
Not enough information is given to
determine the value of x.
20
38. MULTIPLE CHOICE Find the perimeter of the
triangle shown at the right to the nearest tenth
of a centimeter.
★ Challenge
A
¡
C
¡
28.4 cm
31.2 cm
B
¡
D
¡
12 cm
30
30 cm
41.6 cm
VISUAL THINKING In Exercises 39–41, use the diagram below. Each
triangle in the diagram is a 45°-45°-90° triangle. At Stage 0, the legs
of the triangle are each 1 unit long.
1
1
Stage 0
Stage 1
Stage 2
Stage 3
Stage 4
39. Find the exact lengths of the legs of the triangles that are added at each stage.
Leave radicals in the denominators of fractions.
40. Describe the pattern of the lengths in Exercise 39.
EXTRA CHALLENGE
www.mcdougallittell.com
556
41. Find the length of a leg of a triangle added in Stage 8. Explain how you
found your answer.
Chapter 9 Right Triangles and Trigonometry
Page 7 of 7
MIXED REVIEW
42. FINDING A SIDE LENGTH A triangle has one side of 9 inches and another of
14 inches. Describe the possible lengths of the third side. (Review 5.5)
FINDING REFLECTIONS Find the coordinates of the reflection without using
a coordinate plane. (Review 7.2)
43. Q(º1, º2) reflected in the x-axis
44. P(8, 3) reflected in the y-axis
45. A(4, º5) reflected in the y-axis
46. B(0, 10) reflected in the x-axis
DEVELOPING PROOF Name a postulate or theorem that can be used to
prove that the two triangles are similar. (Review 8.5 for 9.5)
47.
48.
49.
6
65
8
9
15
9
10
5
4.5
65
4
INT
10
NE
ER T
Pythagorean Theorem Proofs
THEN
NOW
APPLICATION LINK
www.mcdougallittell.com
AROUND THE SIXTH CENTURY B.C., the Greek mathematician Pythagoras founded a
school for the study of philosophy, mathematics, and science. Many people believe
that an early proof of the Pythagorean Theorem came from this school.
TODAY, the Pythagorean theorem is one of the most famous theorems in geometry.
More than 100 different proofs now exist.
The diagram is based on one drawn by the Hindu mathematician
–skara (1114–1185). The four blue right triangles are congruent.
Bha
a
1. Write an expression in terms of a and b for the combined
areas of the blue triangles. Then write an expression in
terms of a and b for the area of the small red square.
b
2. Use the diagram to show that a2 + b2 = c2. (Hint: This
proof of the Pythagorean Theorem is similar to the one
in Exercise 37 on page 540.)
Chinese manuscript
includes a diagram
that can be used to
prove the theorem.
Nicaraguan stamp
commemorates the
Pythagorean Theorem.
1876
c. 529 B . C .
School of Pythagoras
is founded.
c. A . D . 275
Future U.S. President
Garfield discovers a
proof of the theorem.
1971
9.4 Special Right Triangles
557
Page 1 of 9
9.5
Trigonometric Ratios
What you should learn
GOAL 1 Find the sine, the
cosine, and the tangent of
an acute angle.
GOAL 2 Use trigonometric
ratios to solve real-life
problems, such as estimating
the height of a tree in
Example 6.
Why you should learn it
RE
FE
To solve real-life
problems, such as in finding
the height of a water
slide in Ex. 37.
AL LI
GOAL 1
FINDING TRIGONOMETRIC RATIOS
A trigonometric ratio is a ratio of the lengths of two sides of a right triangle.
The word trigonometry is derived from the ancient Greek language and means
measurement of triangles. The three basic trigonometric ratios are sine, cosine,
and tangent, which are abbreviated as sin, cos, and tan, respectively.
T R I G O N O M E T R I C R AT I O S
Let ¤ABC be a right triangle. The sine, the cosine, and the tangent of the
acute angle ™A are defined as follows.
B
a
side opposite ™A
sin A = = hypotenuse
c
hypotenuse
side adjacent to ™A
hypotenuse
b
c
side opposite ™A
side adjacent to ™A
a
b
side
a opposite
™A
c
cos A = = tan A = = A
b
side adjacent to ™A
C
The value of a trigonometric ratio depends only on the measure of the acute
angle, not on the particular right triangle that is used to compute the value.
EXAMPLE 1
Finding Trigonometric Ratios
Compare the sine, the cosine, and the tangent ratios for ™A in each triangle below.
SOLUTION
By the SSS Similarity Theorem,
the triangles are similar. Their
corresponding sides are in
proportion, which implies that
the trigonometric ratios for ™A
in each triangle are the same.
opposite
hypotenuse
adjacent
cos A = hypotenuse
opposite
tan A = adjacent
sin A = 558
Chapter 9 Right Triangles and Trigonometry
B
17
A
15
B
8
C
8.5
A
Large triangle
Small triangle
8 ≈ 0.4706
17
15
≈ 0.8824
17
4 ≈ 0.4706
8.5
7.
5 ≈ 0.8824
8.5
8 ≈ 0.5333
15
4 ≈ 0.5333
7.5
7.5
4
C
Page 2 of 9
Trigonometric ratios are frequently expressed as decimal approximations.
EXAMPLE 2
INT
STUDENT HELP
NE
ER T
HOMEWORK HELP
Visit our Web site
www.mcdougallittell.com
for extra examples.
Finding Trigonometric Ratios
Find the sine, the cosine, and the tangent
of the indicated angle.
a. ™S
R
13
5
b. ™R
12
T
SOLUTION
S
a. The length of the hypotenuse is 13. For ™S, the length of the opposite side
is 5, and the length of the adjacent side is 12.
opp.
5
13
hyp.
adj.
12
cos S = = ≈ 0.9231
hyp.
13
opp.
5
tan S = = ≈ 0.4167
12
adj.
sin S = = ≈ 0.3846
R
5
opp.
T
13 hyp.
12 adj.
S
b. The length of the hypotenuse is 13. For ™R, the length of the opposite side
is 12, and the length of the adjacent side is 5.
opp.
12
hyp.
13
adj.
5
cos R = = ≈ 0.3846
hyp.
13
opp.
12
tan R = = = 2.4
adj.
5
R
sin R = = ≈ 0.9231
5
adj.
T
13 hyp.
12 opp.
S
..........
You can find trigonometric ratios for 30°, 45°, and 60° by applying what you
know about special right triangles.
EXAMPLE 3
Trigonometric Ratios for 45°
Find the sine, the cosine, and the tangent of 45°.
SOLUTION
STUDENT HELP
Study Tip
The expression sin 45°
means the sine of an
angle whose measure
is 45°.
Begin by sketching a 45°-45°-90° triangle. Because all such
triangles are similar, you can make calculations simple by
choosing 1 as the length of each leg. From Theorem 9.8 on
page 551, it follows that the length of the hypotenuse is 2.
opp.
hyp.
1
2
2
2
adj.
hyp.
1
2
2
2
opp.
adj.
1
1
1
2
hyp.
45
1
sin 45° = = = ≈ 0.7071
cos 45° = = = ≈ 0.7071
tan 45° = = = 1
9.5 Trigonometric Ratios
559
Page 3 of 9
EXAMPLE 4
Trigonometric Ratios for 30°
Find the sine, the cosine, and the tangent of 30°.
SOLUTION
Begin by sketching a 30°-60°-90° triangle. To
make the calculations simple, you can choose 1 as
the length of the shorter leg. From Theorem 9.9 on
page 551, it follows that the length of the longer
leg is 3 and the length of the hypotenuse is 2.
2
1
30
3
opp.
1
hyp.
2
adj.
3
cos 30° = = ≈ 0.8660
hyp.
2
opp.
1
3
tan 30° = = = ≈ 0.5774
adj.
3
3
sin 30° = = = 0.5
EXAMPLE 5
Using a Calculator
You can use a calculator to approximate the sine, the cosine, and the tangent
of 74°. Make sure your calculator is in degree mode. The table shows some
sample keystroke sequences accepted by most calculators.
Sample keystroke sequences
Sample calculator display
Rounded
approximation
74
or
74
0.961261695
0.9613
74
or
74
0.275637355
0.2756
74
or
74
3.487414444
3.4874
..........
STUDENT HELP
Trig Table
For a table of
trigonometric ratios,
see p. 845.
If you look back at Examples 1–5, you will notice that the sine or the cosine
of an acute angle is always less than 1. The reason is that these trigonometric
ratios involve the ratio of a leg of a right triangle to the hypotenuse. The length
of a leg of a right triangle is always less than the length of its hypotenuse, so the
ratio of these lengths is always less than one.
Because the tangent of an acute angle involves the ratio of one leg to another leg,
the tangent of an angle can be less than 1, equal to 1, or greater than 1.
TRIGONOMETRIC IDENTITIES A trigonometric
identity is an equation involving trigonometric
ratios that is true for all acute angles. You are
asked to prove the following identities in
Exercises 47 and 52:
(sin A)2 + (cos A)2 = 1
sin A
cos A
tan A = 560
Chapter 9 Right Triangles and Trigonometry
B
c
A
b
a
C
Page 4 of 9
GOAL 2
USING TRIGONOMETRIC RATIOS IN REAL LIFE
Suppose you stand and look up at a point in the distance, such as the top of the
tree in Example 6. The angle that your line of sight makes with a line drawn
horizontally is called the angle of elevation.
EXAMPLE 6
FOCUS ON
CAREERS
Indirect Measurement
FORESTRY You are measuring the height of a Sitka
spruce tree in Alaska. You stand 45 feet from the
base of the tree. You measure the angle of elevation
from a point on the ground to the top of the tree to
be 59°. To estimate the height of the tree, you can
write a trigonometric ratio that involves the height
h and the known length of 45 feet.
opposite
adjacent
Write ratio.
h
45
Substitute.
tan 59° = tan 59° = 45 tan 59° = h
RE
FE
L
AL I
FORESTRY
45(1.6643) ≈ h
74.9 ≈ h
EXAMPLE 7
L
AL I
CAREER LINK
www.mcdougallittell.com
RE
NE
ER T
59
45 ft
Multiply each side by 45.
Use a calculator or table to find tan 59°.
Simplify.
The tree is about 75 feet tall.
FE
INT
Foresters manage
and protect forests. Their
work can involve measuring
tree heights. Foresters can
use an instrument called a
clinometer to measure the
angle of elevation from a
point on the ground to the
top of a tree.
Estimating a Distance
ESCALATORS The escalator at the Wilshire/Vermont
Metro Rail Station in Los Angeles rises 76 feet at
a 30° angle. To find the distance d a person travels on the
escalator stairs, you can write a trigonometric ratio that
involves the hypotenuse and the known leg length of
76 feet.
opposite
hypotenuse
Write ratio for sine of 30°.
76
d
Substitute.
sin 30° = sin 30° = d sin 30° = 76
d
76 ft
30
Multiply each side by d.
76
sin 30°
Divide each side by sin 30°.
d = 76
0.5
Substitute 0.5 for sin 30°.
d = 152
Simplify.
d = h
A person travels 152 feet on the escalator stairs.
9.5 Trigonometric Ratios
561
Page 5 of 9
GUIDED PRACTICE
In Exercises 1 and 2, use the diagram at the right.
Vocabulary Check
✓
1. Use the diagram to explain what is meant by the
sine, the cosine, and the tangent of ™A.
Concept Check
✓
B
F
37°
A
sin D > sin A because the side lengths of
¤DEF are greater than the side lengths of
¤ABC. Explain why the student is incorrect.
Skill Check
✓
C
2. ERROR ANALYSIS A student says that
In Exercises 3–8, use the diagram shown at the
right to find the trigonometric ratio.
3. sin A
4. cos A
5. tan A
6. sin B
7. cos B
8. tan B
9.
37°
D
E
A
5
3
C
4
ESCALATORS One early escalator built in
1896 rose at an angle of 25°. As shown in the
diagram at the right, the vertical lift was 7 feet.
Estimate the distance d a person traveled on
this escalator.
B
d
7 ft
25
PRACTICE AND APPLICATIONS
STUDENT HELP
Extra Practice
to help you master
skills is on p. 820.
FINDING TRIGONOMETRIC RATIOS Find the sine, the cosine, and the
tangent of the acute angles of the triangle. Express each value as a
decimal rounded to four places.
10.
11.
R
53
28
10
6
A
13.
14.
E
7
STUDENT HELP
24
HOMEWORK HELP
Example 1: Exs. 10–15,
28–36
Example 2: Exs. 10–15,
28–36
Example 3: Exs. 34–36
Example 4: Exs. 34–36
Example 5: Exs. 16–27
Example 6: Exs. 37–42
Example 7: Exs. 37–42
562
12. X
8
2
C
Z
13
S
45
T
B
G
1
J
15.
L
F
5
J
D
5
3
2
25
Y
3
34
H
CALCULATOR Use a calculator to approximate the given value to four
decimal places.
16. sin 48°
17. cos 13°
18. tan 81°
19. sin 27°
20. cos 70°
21. tan 2°
22. sin 78°
23. cos 36°
24. tan 23°
25. cos 63°
26. sin 56°
27. tan 66°
Chapter 9 Right Triangles and Trigonometry
K
Page 6 of 9
USING TRIGONOMETRIC RATIOS Find the value of each variable. Round
decimals to the nearest tenth.
28.
29.
x
30.
34
t
6
4
s
23
36
s
37
r
y
31.
32.
u
65
8
33.
9
w
6
22
v
70
x
y
t
FINDING AREA Find the area of the triangle. Round decimals to the
nearest tenth.
34.
35.
36.
8m
60
45
4 cm
FOCUS ON
APPLICATIONS
37.
38.
RE
FE
L
AL I
30
11 m
12 m
WATER SLIDE The angle of elevation from
the base to the top of a waterslide is about 13°.
The slide extends horizontally about 58.2 meters.
Estimate the height h of the slide.
SURVEYING To find the distance d
from a house on shore to a house on an
island, a surveyor measures from the
house on shore to point B, as shown in
the diagram. An instrument called a
transit is used to find the measure of
™B. Estimate the distance d.
13
h
58.2 m
40 m
B
42
d
WATER SLIDE
Even though riders
on a water slide may travel
at only 20 miles per hour, the
curves on the slide can
make riders feel as though
they are traveling much
faster.
39.
SKI SLOPE Suppose you stand at the top of a ski slope and look down
at the bottom. The angle that your line of sight makes with a line drawn
horizontally is called the angle of depression, as shown below. The vertical
drop is the difference in the elevations of the top and the bottom of the slope.
Find the vertical drop x of the slope in the diagram. Then estimate the
distance d a person skiing would travel on this slope.
elevation
5500 ft
angle of depression
d
elevation
5018 ft
20
x vertical
drop
9.5 Trigonometric Ratios
563
Page 7 of 9
FOCUS ON
APPLICATIONS
40.
Scientists can
measure the depths of craters on the
moon by looking at photos of shadows.
The length of the shadow cast by the
edge of a crater is about 500 meters.
The sun’s angle of elevation is 55°.
Estimate the depth d of the crater.
SCIENCE
FE
RE
55
d
55
500 m
pieces have wheels and a handle so that
the luggage can be pulled along the ground.
Suppose a person’s hand is about 30 inches
from the floor. About how long should the
handle be on the suitcase shown so that
it can roll at a comfortable angle of 45°
with the floor?
LUNAR CRATERS
Because the moon
has no atmosphere to
protect it from being hit by
meteorites, its surface is
pitted with craters. There is
no wind, so a crater can
remain undisturbed for
millions of years––unless
another meteorite crashes
into it.
sun’s ray
LUGGAGE DESIGN Some luggage
41.
L
AL I
CONNECTION
x
26 in.
30 in.
45
BUYING AN AWNING Your family room
42.
sun’s ray
has a sliding-glass door with a southern exposure.
You want to buy an awning for the door that will
be just long enough to keep the sun out when it
is at its highest point in the sky. The angle of
elevation of the sun at this point is 70°, and the
height of the door is 8 feet. About how far should
the overhang extend?
x
8 ft
70
CRITICAL THINKING In Exercises 43 and 44, use the diagram.
43. Write expressions for the sine, the cosine, and the tangent
B
of each acute angle in the triangle.
44.
Writing Use your results from Exercise 43 to explain
c
how the tangent of one acute angle of a right triangle
is related to the tangent of the other acute angle. How
are the sine and the cosine of one acute angle of a right
triangle related to the sine and the cosine of the other
acute angle?
45.
A
b
a
C
TECHNOLOGY Use geometry software to construct a right triangle.
Use your triangle to explore and answer the questions below. Explain
your procedure.
•
•
•
For what angle measure is the tangent of an acute angle equal to 1?
For what angle measures is the tangent of an acute angle greater than 1?
For what angle measures is the tangent of an acute angle less than 1?
Æ
46. ERROR ANALYSIS To find the length of BC
B
in the diagram at the right, a student writes
18
18
tan 55° = . What mistake is the student
BC
making? Show how the student can find BC.
(Hint: Begin by drawing an altitude from
Æ
B to AC.)
564
Chapter 9 Right Triangles and Trigonometry
30°
A
55°
C
Page 8 of 9
B
PROOF Use the diagram of ¤ABC. Complete
47.
c
the proof of the trigonometric identity below.
(sin A)2 + (cos A)2 = 1
a
c
A
a
C
b
b
c
GIVEN sin A = , cos A = PROVE (sin A)2 + (cos A)2 = 1
Statements
a
b
1. sin A = , cos A = c
c
2
2
2
2. a + b = c
a2
b2
3. 2 + 2 = 1
c
c
4.
ac + bc = 1
2
2
5. (sin A)2 + (cos A)2 = 1
Reasons
?
1. ?
2. ?
3. 4. A property of exponents
?
5. DEMONSTRATING A FORMULA Show that (sin A)2 + (cos A)2 = 1 for the
given angle measure.
48. m™A = 30°
52.
Test
Preparation
49. m™A = 45°
50. m™A = 60°
51. m™A = 13°
PROOF Use the diagram in Exercise 47. Write a two-column proof of the
sin A
following trigonometric identity: tan A = .
cos A
53. MULTIPLE CHOICE Use the diagram at the right.
D
Find CD.
A
¡
D
¡
8 cos 25°
8
sin 25°
B
¡
E
¡
8 sin 25°
C
¡
8
8 tan 25°
25
C
8
cos 25°
54. MULTIPLE CHOICE Use the diagram at the
E
B
right. Which expression is not equivalent to AC?
A
¡
D
¡
★ Challenge
EXTRA CHALLENGE
www.mcdougallittell.com
55.
BC sin 70°
BA
tan 20°
B
¡
E
¡
BC cos 20°
C
¡
BC
tan 20°
A
160
C
BA tan 70°
PARADE You are at a parade
looking up at a large balloon floating
directly above the street. You are
60 feet from a point on the street
directly beneath the balloon. To see the
top of the balloon, you look up at an
angle of 53°. To see the bottom of the
balloon, you look up at an angle of 29°.
Estimate the height h of the balloon
to the nearest foot.
h
53
29
60 ft
9.5 Trigonometric Ratios
565
Page 9 of 9
MIXED REVIEW
56. SKETCHING A DILATION ¤PQR is mapped onto ¤P§Q§R§ by a dilation.
In ¤PQR, PQ = 3, QR = 5, and PR = 4. In ¤P§Q§R§, P§Q§ = 6. Sketch the
dilation, identify it as a reduction or an enlargement, and find the scale factor.
Then find the length of Q§R§ and P§R§. (Review 8.7)
57. FINDING LENGTHS Write similarity
N
statements for the three similar triangles in
the diagram. Then find QP and NP. Round
decimals to the nearest tenth. (Review 9.1)
7
M
q
15
P
PYTHAGOREAN THEOREM Find the unknown side length. Simplify answers
that are radicals. Tell whether the side lengths form a Pythagorean triple.
(Review 9.2 for 9.6)
58.
59.
x
60.
65
x
42.9
95
x
50
193
70
QUIZ 2
Self-Test for Lessons 9.4 and 9.5
Sketch the figure that is described. Then find the requested information.
Round decimals to the nearest tenth. (Lesson 9.4)
1. The side length of an equilateral triangle is 4 meters. Find the length of
an altitude of the triangle.
2. The perimeter of a square is 16 inches. Find the length of a diagonal.
3. The side length of an equilateral triangle is 3 inches. Find the area of
the triangle.
Find the value of each variable. Round decimals to the nearest tenth.
(Lesson 9.5)
4.
5.
x
10
40
y
7.
566
y
y
x
25
20
x
62
18
HOT-AIR BALLOON The ground
crew for a hot-air balloon can see the
balloon in the sky at an angle of
elevation of 11°. The pilot radios to the
crew that the hot-air balloon is 950 feet
above the ground. Estimate the horizontal
distance d of the hot-air balloon from the
ground crew. (Lesson 9.5)
Chapter 9 Right Triangles and Trigonometry
6.
Not drawn to scale
ground
crew
11
d
950 f t
Page 1 of 6
9.6
Solving Right Triangles
What you should learn
GOAL 1
Solve a right
triangle.
GOAL 2 Use right triangles
to solve real-life problems,
such as finding the glide
angle and altitude of a
space shuttle in Example 3.
Why you should learn it
RE
FE
To solve real-life problems
such as determining the
correct dimensions of a
wheel-chair ramp in
Exs. 39–41.
AL LI
GOAL 1
SOLVING A RIGHT TRIANGLE
Every right triangle has one right angle, two acute angles, one hypotenuse, and
two legs. To solve a right triangle means to determine the measures of all six
parts. You can solve a right triangle if you know either of the following:
•
•
Two side lengths
One side length and one acute angle measure
As you learned in Lesson 9.5, you can use the side lengths of a right triangle to
find trigonometric ratios for the acute angles of the triangle. As you will see in
this lesson, once you know the sine, the cosine, or the tangent of an acute angle,
you can use a calculator to find the measure of the angle.
In general, for an acute angle A:
if sin A = x, then sinº1 x = m™A.
The expression sin–1 x is read as “the
inverse sine of x.”
if cos A = y, then cosº1 y = m™A.
if tan A = z, then tanº1 z = m™A.
ACTIVITY
Developing
Concepts
1
2
3
Finding Angles in Right Triangles
Carefully draw right ¤ABC with side
lengths of 3 centimeters, 4 centimeters,
and 5 centimeters, as shown.
Use trigonometric ratios to find
the sine, the cosine, and the
tangent of ™A. Express the ratios
in decimal form.
B
5 cm
A
4 cm
3 cm
C
3
In Step 2, you found that sin A = = 0.6. You can use a calculator
5
to find sinº1 0.6. Most calculators use one of the keystroke sequences
below.
sinº1
sinº1
or 0.6
Make sure your calculator is in degree mode. Then use each of
the trigonometric ratios you found in Step 2 to approximate the
measure of ™A to the nearest tenth of a degree.
0.6
4
Use a protractor to measure ™A. How does the measured value
compare with your calculated values?
9.6 Solving Right Triangles
567
Page 2 of 6
Solving a Right Triangle
EXAMPLE 1
C
Solve the right triangle. Round decimals
to the nearest tenth.
3
2
c
B
A
SOLUTION
Begin by using the Pythagorean Theorem to find the length of the hypotenuse.
(hypotenuse)2 = (leg)2 + (leg)2
2
2
c =3 +2
Pythagorean Theorem
2
Substitute.
2
Simplify.
c = 13
Find the positive square root.
c ≈ 3.6
Use a calculator to approximate.
c = 13
Then use a calculator to find the measure of ™B:
2
3
≈ 33.7°
Finally, because ™A and ™B are complements, you can write
m™A = 90° º m™B ≈ 90° º 33.7° = 56.3°.
The side lengths of the triangle are 2, 3, and 1
3 , or about 3.6. The triangle has
one right angle and two acute angles whose measures are about 33.7° and 56.3°.
EXAMPLE 2
Solving a Right Triangle
Solve the right triangle. Round decimals
to the nearest tenth.
H
g
25
13
Study Tip
There are other ways to
find the side lengths in
Examples 1 and 2. For
instance, in Example 2,
you can use a trigonometric ratio to find one
side length, and then
use the Pythagorean
Theorem to find the other
side length.
h
G
SOLUTION
STUDENT HELP
J
Use trigonometric ratios to find the values of g and h.
opp.
hyp.
sin H = h
13
sin 25° = adj.
hyp.
g
cos 25° = 13
cos H = 13 sin 25° = h
13 cos 25° = g
13(0.4226) ≈ h
13(0.9063) ≈ g
5.5 ≈ h
11.8 ≈ g
Because ™H and ™G are complements, you can write
m™G = 90° º m™H = 90° º 25° = 65°.
568
The side lengths of the triangle are about 5.5, 11.8, and 13. The triangle has
one right angle and two acute angles whose measures are 65° and 25°.
Chapter 9 Right Triangles and Trigonometry
Page 3 of 6
GOAL 2
USING RIGHT TRIANGLES IN REAL LIFE
EXAMPLE 3
FOCUS ON
CAREERS
Solving a Right Triangle
SPACE SHUTTLE During its approach to
Earth, the space shuttle’s glide angle changes.
Not drawn
to scale
UNITE
D STA
TES
a. When the shuttle’s altitude is about
15.7 miles, its horizontal distance to
the runway is about 59 miles. What
is its glide angle? Round your answer
to the nearest tenth.
altitude
glide angle
b. When the space shuttle is 5 miles from
RE
FE
L
AL I
the runway, its glide angle is about 19°.
Find the shuttle’s altitude at this point in
its descent. Round your answer to the
nearest tenth.
runway
ASTRONAUT
Some astronauts are
pilots who are qualified to fly
the space shuttle. Some
shuttle astronauts are
mission specialists whose
responsibilities include
conducting scientific
experiments in space. All
astronauts need to have a
strong background in
science and mathematics.
INT
distance
to runway
SOLUTION
a. Sketch a right triangle to model the situation.
Let x° = the measure of the shuttle’s glide angle.
You can use the tangent ratio and a calculator to
find the approximate value of x.
opp.
tan x° = adj.
15.7
59
tan x° = NE
ER T
CAREER LINK
www.mcdougallittell.com
x=
x
59 mi
15.7 mi
Substitute.
15.7
59
Use a calculator to
15.7
find tanº1 }} .
59 x ≈ 14.9
When the space shuttle’s altitude is about 15.7 miles, the glide angle is
about 14.9°.
b. Sketch a right triangle to model the situation.
Let h = the altitude of the shuttle. You can
use the tangent ratio and a calculator to find
the approximate value of h.
opp.
tan 19° = adj.
h
5
h
0.3443 ≈ 5
tan 19° = INT
STUDENT HELP
NE
ER T
1.7 ≈ h
HOMEWORK HELP
Visit our Web site
www.mcdougallittell.com
for extra examples.
h
19
5 mi
Substitute.
Use a calculator.
Multiply each side by 5.
The shuttle’s altitude is about 1.7 miles.
9.6 Solving Right Triangles
569
Page 4 of 6
GUIDED PRACTICE
✓
Concept Check ✓
Vocabulary Check
1. Explain what is meant by solving a right triangle.
Tell whether the statement is true or false.
2. You can solve a right triangle if you are given the lengths of any two sides.
3. You can solve a right triangle if you know only the measure of one acute angle.
Skill Check
✓
CALCULATOR In Exercises 4–7, ™A is an acute angle. Use a calculator to
approximate the measure of ™A to the nearest tenth of a degree.
4. tan A = 0.7
5. tan A = 5.4
6. sin A = 0.9
7. cos A = 0.1
Solve the right triangle. Round decimals to the nearest tenth.
8.
9.
A
33
c
C
56
109
B
10. X
E
D
y
d
91
Z
x
4
60
Y
F
PRACTICE AND APPLICATIONS
STUDENT HELP
Extra Practice
to help you master
skills is on p. 820.
q
FINDING MEASUREMENTS Use the diagram to find
the indicated measurement. Round your answer to
the nearest tenth.
11. QS
12. m™Q
48
13. m™S
T
S
55
CALCULATOR In Exercises 14–21, ™A is an acute angle. Use a
calculator to approximate the measure of ™A to the nearest tenth
of a degree.
14. tan A = 0.5
15. tan A = 1.0
16. sin A = 0.5
17. sin A = 0.35
18. cos A = 0.15
19. cos A = 0.64
20. tan A = 2.2
21. sin A = 0.11
SOLVING RIGHT TRIANGLES Solve the right triangle. Round decimals to
the nearest tenth.
22.
23.
A
24.
F
G
7
7
20
2
21
C
STUDENT HELP
D
B
E
J
6
H
12.5
S
HOMEWORK HELP
Example 1: Exs. 11–27,
34–37
Example 2: Exs. 28–33
Example 3: Exs. 38–41
25.
27.
P
8
K
570
26.
M
13.6
9.2
L
Chapter 9 Right Triangles and Trigonometry
N
4
q
R
6
T
Page 5 of 6
SOLVING RIGHT TRIANGLES Solve the right triangle. Round decimals to
the nearest tenth.
28.
29.
P
30.
X
U
q
4.5
R
26
p
q
31.
S
t
s
20
12
T
Y
32.
D
52
z
33.
8.5
Z
x
L
C
A
51
e
a
5
56
3
4
E
M
m
34
c
B
F
d
L
K
NATIONAL AQUARIUM Use the diagram
of one of the triangular windowpanes at the
National Aquarium in Baltimore, Maryland,
to find the indicated value.
?
34. tan B ≈ B
?
35. m™B ≈ 36 in.
?
36. AB ≈ A
69 in.
C
?
37. sin A ≈ ..
FOCUS ON
38.
APPLICATIONS
RE
FE
L
AL I
s
HIKING You are hiking up a mountain
peak. You begin hiking at a trailhead whose
elevation is about 9400 feet. The trail ends
near the summit at 14,255 feet. The horizontal
distance between these two points is about
17,625 feet. Estimate the angle of elevation
from the trailhead to the summit.
summit
14,255 ft
trailhead
9400 ft
horizontal distance 17,625 ft
RAMPS In Exercises 39–41, use the
information about wheelchair ramps.
BENCHMARKS
If you hike to the top
of a mountain you may find a
brass plate called a
benchmark. A benchmark
gives an official elevation for
the point that can be used by
surveyors as a reference for
surveying elevations of other
landmarks.
The Uniform Federal Accessibility
Standards specify that the ramp angle
used for a wheelchair ramp must be less
than or equal to 4.76°.
length of ramp
ramp angle
vertical
rise
horizontal distance
39. The length of one ramp is 20 feet. The vertical rise is 17 inches.
Estimate the ramp’s horizontal distance and its ramp angle.
40. You want to build a ramp with a vertical rise of 8 inches. You want to
minimize the horizontal distance taken up by the ramp. Draw a sketch
showing the approximate dimensions of your ramp.
41.
Writing
Measure the horizontal distance and the vertical rise of a ramp
near your home or school. Find the ramp angle. Does the ramp meet the
specifications described above? Explain.
9.6 Solving Right Triangles
571
Page 6 of 6
Test
Preparation
MULTI-STEP PROBLEM In Exercises 42–45,
use the diagram and the information below.
The horizontal part of a step is called the tread.
The vertical part is called the riser. The ratio of
the riser length to the tread length affects the safety
of a staircase. Traditionally, builders have used a
riser-to-tread ratio of about 814 inches : 9 inches.
A newly recommended ratio is 7 inches : 11 inches.
tread
riser
x
42. Find the value of x for stairs built using the new riser-to-tread ratio.
43. Find the value of x for stairs built using the old riser-to-tread ratio.
44. Suppose you want to build a stairway that is less steep than either of the ones
in Exercises 42 and 43. Give an example of a riser-to-tread ratio that you
could use. Find the value of x for your stairway.
45.
★ Challenge
Writing Explain how the riser-to-tread ratio that is used for a stairway
could affect the safety of the stairway.
C
PROOF Write a proof.
46.
GIVEN ™A and ™B are acute angles.
EXTRA CHALLENGE
www.mcdougallittell.com
a
sin A
b
sin B
PROVE = a
b
c
A
B
Æ
( Hint: Draw an altitude from C to AB. Label it h.)
MIXED REVIEW
USING VECTORS Write the component form
of the vector. (Review 7.4 for 9.7)
Æ„
Æ„
47. AB
48. AC
Æ„
Æ„
49. DE
Æ„
51. FH
Æ„
B
D
1
50. FG
52. JK
y
C
A
x
1
F
E
K
J
G
H
SOLVING PROPORTIONS Solve the proportion. (Review 8.1)
x
5
53. = 30
6
7
49
54. = 16
y
g
3
55. = 10
42
7
84
56. = 18
k
m
7
57. = 2
1
4
8
58. = 11
t
CLASSIFYING TRIANGLES Decide whether the numbers can represent the
side lengths of a triangle. If they can, classify the triangle as right, acute, or
obtuse. (Review 9.3)
572
59. 18, 14, 2
60. 60, 228, 220
61. 8.5, 7.7, 3.6
62. 250, 263, 80
63. 113, 15, 112
64. 15, 75, 59
Chapter 9 Right Triangles and Trigonometry
Page 1 of 8
9.7
Vectors
What you should learn
GOAL 1 Find the magnitude
and the direction of a vector.
GOAL 2
Add vectors.
Why you should learn it
RE
FE
To solve real-life
problems, such as
describing the velocity
of a skydiver in
Exs. 41–45.
AL LI
GOAL 1
FINDING THE MAGNITUDE OF A VECTOR
As defined in Lesson 7.4, a vector is a quantity that has both magnitude and
direction. In this lesson, you will learn how to find the magnitude of a vector and
the direction of a vector. You will also learn how to add vectors.
Æ„
The magnitude of a vector AB is the distance
from the initial point A to the terminal point B, and
Æ„
is written |AB |. If a vector is drawn in a coordinate
plane, you can use the Distance Formula to find its
magnitude.
y
B(x 2, y 2)
A(x 1, y 1)
2
2
x
(
y2
ºy
|AB | = (x
2º
1)+
1)
Æ„
EXAMPLE 1
x
Finding the Magnitude of a Vector
Æ„
Æ„
Points P and Q are the initial and terminal points of the vector PQ . Draw PQ in a
coordinate plane. Write the component form of the vector and find its magnitude.
a. P(0, 0), Q(º6, 3)
b. P(0, 2), Q(5, 4)
c. P(3, 4), Q(º2, º1)
SOLUTION
a. Component form = ⟨x 2 º x1, y2 º y1⟩
y
Æ„
PQ = ⟨º6 º 0, 3 º 0⟩
= ⟨º6, 3⟩
œ
2
Use the Distance Formula to find the magnitude.
P
1
|PQ | = (º
6
º0
)
+(3
º0
) = 45 ≈ 6.7
Æ„
STUDENT HELP
Look Back
For help with the
component form of a
vector, see p. 423.
2
2
b. Component form = ⟨x 2 º x1, y2 º y1⟩
y
œ
Æ„
PQ = ⟨5 º 0, 4 º 2⟩
= ⟨5, 2⟩
P
Use the Distance Formula to find the magnitude.
1
1
|PQ | = (5
º0
)2
+(4
º
2
)2 = 29 ≈ 5.4
Æ„
x
c. Component form = ⟨x2 º x1, y2 º y1⟩
y
P
Æ„
PQ = ⟨º2 º 3, º1 º 4⟩
2
= ⟨º5, º5⟩
Use the Distance Formula to find the magnitude.
|PQ | = (º
2
º3
)
+(º
1
º4
) = 50 ≈ 7.1
Æ„
2
x
2
1
x
œ
9.7 Vectors
573
Page 2 of 8
FOCUS ON
APPLICATIONS
The direction of a vector is determined by the angle it makes with a
horizontal line. In real-life applications, the direction angle is described relative
to the directions north, east, south, and west. In a coordinate plane, the x-axis
represents an east-west line. The y-axis represents a north-south line.
EXAMPLE 2
Describing the Direction of a Vector
Æ„
RE
FE
L
AL I
NAVIGATION
One of the most
common vector quantities
in real life is the velocity of
a moving object. A velocity
vector is used in navigation
to describe both the speed
and the direction of a
moving object.
The vector AB describes the velocity of a moving
ship. The scale on each axis is in miles per hour.
N
y
B
a. Find the speed of the ship.
b. Find the direction it is traveling
relative to east.
5 A
W
5
x
E
S
SOLUTION
Æ„
a. The magnitude of the vector AB represents the
ship’s speed. Use the Distance Formula.
|AB | = (2
5
º5
)2
+(20
º5
)2
Æ„
= 202+
152
= 25
STUDENT HELP
Look Back
For help with using
trigonometric ratios to
find angle measures, see
pp. 567 and 568.
The speed of the ship is 25 miles per hour.
b. The tangent of the angle formed by the vector
15
and a line drawn parallel to the x-axis is ,
20
N
y
B
or 0.75. Use a calculator to find the angle measure.
0.75
15
5 A
≈ 36.9°
W
?
20
5
The ship is traveling in a direction about 37°
north of east.
..........
x
E
S
Two vectors are equal if they have the same magnitude and direction. They do
not have to have the same initial and terminal points. Two vectors are parallel
if they have the same or opposite directions.
EXAMPLE 3
Identifying Equal and Parallel Vectors
In the diagram, these vectors have the same
Æ„ Æ„ Æ„
direction: AB , CD , EF .
y
G
C
A
Æ„ Æ„
These vectors are equal: AB , CD .
H
D
Æ„ Æ„ Æ„ Æ„
E
These vectors are parallel: AB , CD , EF , HG .
B
1
J
1
x
K
F
574
Chapter 9 Right Triangles and Trigonometry
Page 3 of 8
GOAL 2
Study Tip
A single letter with an
arrow over it, such as u„,
can be used to denote
a vector.
Two vectors can be added to form a new vector. To
v geometrically, place the initial point
add „
u and „
„
u , (or place the initial
of v on the terminal point of „
v ). The sum is
point of „
u on the terminal point of „
the vector that joins the initial point of the first
vector and the terminal point of the second vector .
y
„
u
„
„
v
„
u v
„
v
This method of adding vectors is often called the
parallelogram rule because the sum vector is the
diagonal of a parallelogram. You can also add
vectors algebraically.
„
u
x
A D D I N G V E C TO R S
SUM OF TWO VECTORS
„
„
„ „
The sum of u
= ⟨a1, b1⟩ and v
= ⟨a2, b2⟩ is u
+ v = ⟨a1 + a2, b1 + b2⟩.
A D D I N G V E C TO R S
EXAMPLE 4
Finding the Sum of Two Vectors
Let „
u = ⟨3, 5⟩ and „
v = ⟨º6, º1⟩. To find the
v , add the horizontal components
sum vector „
u +„
v.
and add the vertical components of „
u and „
y
„
u + v = ⟨3 + (º6), 5 + (º1)⟩
„
„
„
u
„
u v
= ⟨º3, 4⟩
„
v
„
u
2
EXAMPLE 5
x
1
„
v
Velocity of a Jet
FE
L
AL I
AVIATION A jet is flying northeast at
about 707 miles per hour. Its velocity
is represented by the vector „
v = ⟨500, 500⟩.
RE
STUDENT HELP
ADDING VECTORS
The jet encounters a wind blowing from the
west at 100 miles per hour. The wind velocity
is represented by „
u = ⟨100, 0⟩. The jet’s new
velocity vector „
s is the sum of its original
velocity vector and the wind’s velocity vector.
N
„
u
„
v
„
s
100
100
E
„
s=„
v+„
u
= ⟨500 + 100, 500 + 0⟩
= ⟨600, 500⟩
Τhe magnitude of the sum vector „
s represents the new speed of the jet.
New speed =|s„| = (6
00
º0
)2
+(500
º0
)2 ≈ 781 mi/h
9.7 Vectors
575
Page 4 of 8
GUIDED PRACTICE
✓
Concept Check ✓
Vocabulary Check
1. What is meant by the magnitude of a vector and the direction of a vector?
In Exercises 2–4, use the diagram.
A y
2. Write the component form of each vector.
B
3. Identify any parallel vectors.
Æ„
V
Æ„
4. Vectors PQ and ST are equal vectors.
✓
2
M
P
Æ„
Skill Check
œ
1
Although ST is not shown, the coordinates
of its initial point are (º1, º1). Give the
coordinates of its terminal point.
x
U
N
Write the vector in component form. Find the magnitude of the vector.
Round your answer to the nearest tenth.
5.
y
6.
B
7.
y
y
M
P
1
1
2
A
1
œ
1 x
1
x
N
x
8. Use the vector in Exercise 5. Find the direction of the vector relative to east.
9. Find the sum of the vectors in Exercises 5 and 6.
PRACTICE AND APPLICATIONS
STUDENT HELP
Extra Practice
to help you master
skills is on p. 820.
FINDING MAGNITUDE Write the vector in component form. Find the
magnitude of the vector. Round your answer to the nearest tenth.
10.
11.
y
12.
y
K
y
1
1 x
S
1
1
R
1
E
2
x
x
J
F
Æ„
STUDENT HELP
HOMEWORK HELP
Example 1:
Example 2:
Example 3:
Example 4:
Example 5:
576
Exs. 10–20
Exs. 21–24
Exs. 25–29
Exs. 31–40
Exs. 41–45
FINDING MAGNITUDE Draw vector PQ in a coordinate plane. Write the
component form of the vector and find its magnitude. Round your answer
to the nearest tenth.
13. P(0, 0), Q(2, 7)
14. P(5, 1), Q(2, 6)
15. P(º3, 2), Q(7, 6)
16. P(º4, º3), Q(2, º7)
17. P(5, 0), Q(º1, º4)
18. P(6, 3), Q(º2, 1)
19. P(º6, 0), Q(º5, º4)
20. P(0, 5), Q(3, 5)
Chapter 9 Right Triangles and Trigonometry
Page 5 of 8
NAVIGATION The given vector represents the velocity of a ship at sea.
Find the ship’s speed, rounded to the nearest mile per hour. Then find the
direction the ship is traveling relative to the given direction.
21. Find direction relative to east.
22. Find direction relative to east.
y
y
10
A
T
x
20
E
S
10
x
10
E
B
23. Find direction relative to west.
24. Find direction relative to west.
y
M
y
10
W
L
W
O
x
10
10
10
x
P
PARALLEL AND EQUAL VECTORS
In Exercises 25–28, use the diagram
shown at the right.
25. Which vectors are parallel?
26. Which vectors have the same direction?
y
F
D
H
1
E
C
A
G
x
1
K
27. Which vectors are equal?
J
B
28. Name two vectors that have the same
magnitude but different directions.
FOCUS ON
APPLICATIONS
TUG-OF-WAR GAME In Exercises 29 and 30, use the information below.
The forces applied in a game of tug-of-war can be represented by vectors. The
magnitude of the vector represents the amount of force with which the rope is
pulled. The direction of the vector represents the direction of the pull. The
diagrams below show the forces applied in two different rounds of tug-of-war.
Round 1
Round 2
Team A
A
RE
FE
L
AL I
TUG-OF-WAR
In the game
tug-of-war, two teams pull
on opposite ends of a rope.
The team that succeeds in
pulling a member of the
other team across a center
line wins.
Team B
C
B
Team A
A
center line
Æ
„
Team B
C
B
center line
Æ„
29. In Round 2, are CA and CB parallel vectors? Are they equal vectors?
30. In which round was the outcome a tie? How do you know? Describe the
outcome in the other round. Explain your reasoning.
9.7 Vectors
577
Page 6 of 8
„
„
PARALLELOGRAM RULE Copy the vectors u
and v
. Write the
„
„
component form of each vector. Then find the sum u
+v
and draw
„
„
the vector u + v .
31.
32.
y
y
„
v
„
v
„
1
u
u
1
x
1
33.
1
„
34.
y
x
y
1
„
u
1
„
v
„
x
„
v
1
u
1
x
„
„
„
ADDING VECTORS Let u
= ⟨7, 3⟩, v
= ⟨1, 4⟩, w
= ⟨3, 7⟩, and z„ = ⟨º3, º7⟩.
Find the given sum.
FOCUS ON
APPLICATIONS
35. „
v+„
w
36. „
u +„
v
37. „
u+„
w
38. „
v+„
z
39. „
u+„
z
40. „
w+„
z
SKYDIVING In Exercises 41–45, use the information and diagram below.
A skydiver is falling at a constant downward velocity of 120 miles per hour. In
the diagram, vector „
u represents the skydiver’s velocity. A steady breeze pushes
the skydiver to the east at 40 miles per hour. Vector „
v represents the wind
velocity. The scales on the axes of the graph are in miles per hour.
41. Write the vectors „
u and „
v in
UP
component form.
42. Let „
s =„
u +„
v . Copy the diagram
RE
FE
L
AL I
and draw vector „
s.
SKYDIVING
A skydiver who has
not yet opened his or her
parachute is in free fall.
During free fall, the skydiver
accelerates at first. Air
resistance eventually stops
this acceleration, and the
skydiver falls at terminal
velocity.
43. Find the magnitude of „
s . What
information does the magnitude give
you about the skydiver’s fall?
u
44. If there were no wind, the skydiver
would fall in a path that was straight
down. At what angle to the ground is
the path of the skydiver when the
skydiver is affected by the 40 mile
per hour wind from the west?
45. Suppose the skydiver was blown to
the west at 30 miles per hour. Sketch
a new diagram and find the skydiver’s
new velocity.
578
„
Chapter 9 Right Triangles and Trigonometry
„
v
10
W
10
DOWN
E
Page 7 of 8
46.
47.
Writing
Write the component form of a vector with the same magnitude as
JK = ⟨1, 3⟩ but a different direction. Explain how you found the vector.
Æ„
LOGICAL REASONING Let vector „
u = ⟨r, s⟩. Suppose the horizontal and
the vertical components of „
u are multiplied by a constant k. The resulting
„
u and „
v
vector is v = ⟨kr, ks⟩. How are the magnitudes and the directions of „
related when k is positive? when k is negative? Justify your answers.
Test
Preparation
48. MULTI-STEP PROBLEM A motorboat heads due east across a river at a
speed of 10 miles per hour. Vector „
u = ⟨10, 0⟩ represents the velocity of the
motorboat. The current of the river is flowing due north at a speed of 2 miles
per hour. Vector „
v = ⟨0, 2⟩ represents the velocity of the current.
a. Let „
s =„
u +„
v . Draw the vectors „
u, „
v,
N
and „
s in a coordinate plane.
b. Find the speed and the direction of the
motorboat as it is affected by the current.
c. Suppose the speed of the motorboat is
greater than 10 miles per hour, and the
speed of the current is less than 2 miles
per hour. Describe one possible set of
v that could represent the
vectors „
u and „
velocity of the motorboat and the
velocity of the current. Write and solve a
word problem that can be solved by
finding the sum of the two vectors.
★ Challenge
„
u
10 mi/h
„
v
current
2 mi/h
BUMPER CARS In Exercises 49–52, use the information below.
As shown in the diagram below, a bumper car moves from point A to point B to
point C and back to point A. The car follows the path shown by the vectors. The
magnitude of each vector represents the distance traveled by the car from the
initial point to the terminal point.
C
60 ft
B
A
54 ft
Æ„
Æ„
49. Find the sum of AB and BC . Write the sum vector in component form.
Æ„
50. Add vector CA to the sum vector from Exercise 49.
EXTRA CHALLENGE
www.mcdougallittell.com
51. Find the total distance traveled by the car.
52. Compare your answers to Exercises 50 and 51. Why are they different?
9.7 Vectors
579
Page 8 of 8
MIXED REVIEW
53.
PROOF Use the information and the
diagram to write a proof. (Review 4.5)
D
B
E
GIVEN ™D and ™E are right angles;
Æ
Æ
¤ABC is equilateral; DE ∞ AC
A
Æ
PROVE B is the midpoint of DE .
C
xy USING ALGEBRA Find the values of x and y. (Review 4.6)
54.
55.
56.
x
y
y
x
x
y
xy USING ALGEBRA Find the product. (Skills Review, p. 798, for 10.1)
57. (x + 1)2
58. (x + 7)2
59. (x + 11)2
QUIZ 3
60. (7 + x)2
Self-Test for Lessons 9.6 and 9.7
Solve the right triangle. Round decimals to the nearest tenth. (Lesson 9.6)
1.
2.
A
46
b
a
B
12
4. P
75
q
y
E
8
6
q
N
6. J
L
K
3
L
f
q
F
M
Z
5.
p
R
40
45
X
m
16
z
25
C
q
3.
Y
7.6
12.4
G
Æ„
Draw vector PQ in a coordinate plane. Write the component form of the
vector and find its magnitude. Round your answer to the nearest tenth.
(Lesson 9.7)
7. P(3, 4), Q(º2, 3)
8. P(º2, 2), Q(4, º3)
9. P(0, º1), Q(3, 4)
Æ„
10. P(2, 6), Q(º5, º5)
Æ„
11. Vector ST = ⟨3, 8⟩. Draw ST in a coordinate plane and find its direction
relative to east. (Lesson 9.7)
„
„
„
Let u
= ⟨0, º5⟩, v
= ⟨4, 7⟩, w
= ⟨º2, º3⟩, and z„= ⟨2, 6⟩. Find the given sum.
(Lesson 9.7)
580
12. „
u +„
v
13. „
v +„
w
14. „
u+„
w
15. „
u+„
z
16. „
v +„
z
17. „
w+„
z
Chapter 9 Right Triangles and Trigonometry
Page 1 of 5
CHAPTER
9
Chapter Summary
WHAT did you learn?
WHY did you learn it?
Solve problems involving similar right triangles
formed by the altitude drawn to the hypotenuse
of a right triangle. (9.1)
Find a height in a real-life structure, such as the
height of a triangular roof. (p. 528)
Use the Pythagorean Theorem. (9.2)
Solve real-life problems, such as finding the length
of a skywalk support beam. (p. 537)
Use the Converse of the Pythagorean Theorem.
Use in construction methods, such as verifying
whether a foundation is rectangular. (p. 545)
(9.3)
Use side lengths to classify triangles by their
angle measures. (9.3)
Write proofs about triangles. (p. 547)
Find side lengths of special right triangles. (9.4)
Solve real-life problems, such as finding the height
of a loading platform. (p. 553)
Find trigonometric ratios of an acute angle.
Measure distances indirectly, such as the depth of a
crater on the moon. (p. 564)
(9.5)
Solve a right triangle. (9.6)
Solve real-life problems, such as finding the glide
angle and altitude of the space shuttle. (p. 569)
Find the magnitude and the direction of a vector.
Describe physical quantities, such as the speed and
direction of a ship. (p. 574)
(9.7)
Find the sum of two vectors. (9.7)
Model real-life motion, such as the path of a
skydiver. (p. 578)
How does Chapter 9 fit into the BIGGER PICTURE of geometry?
In this chapter, you studied two of the most important theorems in mathematics—
the Pythagorean Theorem and its converse. You were also introduced to a branch of
mathematics called trigonometry. Properties of right triangles allow you to estimate
distances and angle measures that cannot be measured directly. These properties are
important tools in areas such as surveying, construction, and navigation.
STUDY STRATEGY
What did you learn
about right triangles?
Your lists about what you knew
and what you expected to learn
about right triangles, following the
study strategy on page 526, may
resemble this one.
What I Already Know About Rig
ht
1. Have a right angle.
2. Perpendicular sides
are legs.
3. Longest side is the
hypotenuse.
leg
Triangles
hypotenuse
leg
581
Page 2 of 5
CHAPTER
9
Chapter Review
• Pythagorean triple, p. 536
• special right triangles, p. 551
• trigonometric ratio, p. 558
• sine, p. 558
9.1
• cosine, p. 558
• tangent, p. 558
• angle of elevation, p. 561
• solve a right triangle, p. 567
• magnitude of a vector, p. 573
• direction of a vector, p. 574
• equal vectors, p. 574
• parallel vectors, p. 574
• sum of two vectors, p. 575
Examples on
pp. 528–530
SIMILAR RIGHT TRIANGLES
C
EXAMPLES
DB
CB
CB
AB
¤ACB ~ ¤CDB, so = . CB is the geometric
mean of DB and AB.
A
D
B
AD
AC
¤ADC ~ ¤ACB, so = . AC is the geometric mean of AD and AB.
AC
AB
DA
DC
DC
DB
¤CDB ~ ¤ADC, so = . DC is the geometric mean of DA and DB.
Find the value of each variable.
1.
2.
y
6
y
9
36
z
25
x
9.2
3.
9
27
y
x
x
Examples on
pp. 536–537
THE PYTHAGOREAN THEOREM
EXAMPLE You can use the Pythagorean Theorem to find the value of r.
172 = r 2 + 152, or 289 = r 2 + 225. Then 64 = r 2, so r = 8.
The side lengths 8, 15, and 17 form a Pythagorean triple because they
are integers.
17
15
The variables r and s represent the lengths of the legs of a right triangle,
and t represents the length of the hypotenuse. Find the unknown value.
Then tell whether the lengths form a Pythagorean triple.
4. r = 12, s = 16
582
5. r = 8, t = 12
Chapter 9 Right Triangles and Trigonometry
6. s = 16, t = 34
7. r = 4, s = 6
r
Page 3 of 5
9.3
Examples on
pp. 543–545
THE CONVERSE OF THE PYTHAGOREAN THEOREM
EXAMPLES You can use side lengths to classify a triangle by its angle measures.
Let a, b, and c represent the side lengths of a triangle, with c as the length of the
longest side.
If c2 = a2 + b2, the triangle is a right triangle:
82 = (27)2 + 62, so 27, 6, and 8 are
the side lengths of a right triangle.
If c2 < a2 + b2, the triangle is an acute triangle:
122 < 82+ 92 , so 8, 9, and 12 are the side
lengths of an acute triangle.
If c2 > a2 + b2, the triangle is an obtuse triangle:
82 > 52 + 62, so 5, 6, and 8 are the
side lengths of an obtuse triangle.
Decide whether the numbers can represent the side lengths of a triangle. If
they can, classify the triangle as acute, right, or obtuse.
8. 6, 7, 10
9.4
9. 9, 40, 41
11. 3, 45
, 9
10. 8, 12, 20
Examples on
pp. 551–553
SPECIAL RIGHT TRIANGLES
EXAMPLES Triangles whose angle measures are 45°-45°-90° or 30°-60°-90° are
called special right triangles.
45
62
6
45°-45°-90° triangle
hypotenuse = 2 • leg
8
60
30
43
45
4
30°-60°-90° triangle
hypotenuse = 2 • shorter leg
longer leg = 3 • shorter leg
6
12. An isosceles right triangle has legs of length 32
. Find the length of the
hypotenuse.
13. A diagonal of a square is 6 inches long. Find its perimeter and its area.
14. A 30°-60°-90° triangle has a hypotenuse of length 12 inches. What are the
lengths of the legs?
15. An equilateral triangle has sides of length 18 centimeters. Find the length of
an altitude of the triangle. Then find the area of the triangle.
9.5
Examples on
pp. 558–561
TRIGONOMETRIC RATIOS
EXAMPLE
A trigonometric ratio is a ratio of the lengths of two sides of a
Y
right triangle.
opp.
20
sin X = = hyp.
29
29
adj.
21
cos X = = hyp.
29
20
opp.
tan X = = 21
adj.
X
21
20
Z
Chapter Review
583
Page 4 of 5
9.5 continued
Find the sine, the cosine, and the tangent of the acute angles of the
triangle. Express each value as a decimal rounded to four places.
16.
L
11
61
60
J
17.
18. B
M
C
7
35
12
K
42
9
37
P
N
A
9.6
SOLVING RIGHT TRIANGLES
Examples on
pp. 568–569
To solve ¤ABC, begin by using the Pythagorean
Theorem to find the length of the hypotenuse.
EXAMPLE
B
c2 = 102 + 152 = 325. So, c = 325 = 513.
Then find m™A and m™B.
A
c
10
15
C
10
2
tan A = = . Use a calculator to find that m™A ≈ 33.7°.
15
3
Then m™B = 90° º m™A ≈ 90° º 33.7° = 56.3°.
Solve the right triangle. Round decimals to the nearest tenth.
19.
20.
Z
12
21.
E
d
f
x
T
s
8
15
S
50
8
X
9.7
20
D
Y
R
F
Examples on
pp. 573–575
VECTORS
EXAMPLES You can use the Distance Formula to find
Æ„
the magnitude of PQ .
y
œ(8, 10)
|PQ | = (8
º2
)2+
(10
º2
)2 = 62+
82 = 100 = 10
Æ„
To add vectors, find the sum of their horizontal components
and the sum of their vertical components.
Æ„
Æ
„
PQ + OT = ⟨6, 8⟩ + ⟨8, º2⟩ = ⟨6 + 8, 8 + (º2)⟩ = ⟨14, 6⟩
2
P(2, 2)
O
10
T(8, 2)
Æ„
Draw vector PQ in a coordinate plane. Write the component form of the
vector and find its magnitude. Round decimals to the nearest tenth.
22. P(2, 3), Q(1, º1)
23. P(º6, 3), Q(6, º2)
24. P(º2, 0), Q(1, 2)
25. Let u = ⟨1, 2⟩ and v = ⟨13, 7⟩. Find u + v . Find the magnitude of the sum
„
„
„
„
vector and its direction relative to east.
584
Chapter 9 Right Triangles and Trigonometry
x
Page 5 of 5
Chapter Test
CHAPTER
9
Use the diagram at the right to match the angle or segment with its
measure. (Some measures are rounded to two decimal places.)
Æ
1. AB
A. 5.33
Æ
2. BC
A
B. 36.87°
Æ
4
3. AD
C. 5
4. ™BAC
D. 53.13°
5. ™CAD
E. 6.67
3
C
B
D
6. Refer to the diagram above. Complete the following statement:
? ~ ¤
? .
¤ABC ~ ¤
W
7. Classify quadrilateral WXYZ in the diagram at the right. Explain your
reasoning.
15
Z
8. The vertices of ¤PQR are P(º2, 3), Q(3, 1), and R(0, º3). Decide
X
8
17
whether ¤PQR is right, acute, or obtuse.
6
Y
? , and 113 form a Pythagorean triple.
9. Complete the following statement: 15, 10. The measure of one angle of a rhombus is 60°. The perimeter of the rhombus
is 24 inches. Sketch the rhombus and give its side lengths. Then find its area.
Solve the right triangle. Round decimals to the nearest tenth.
11.
12.
K
13.
E
P
12
9
D
30
J
6
4
25
F
L
q
R
Æ„
14. L = (3, 7) and M = (7, 4) are the initial and the terminal points of LM . Draw
Æ„
LM in a coordinate plane. Write the component form of the vector. Then find
its magnitude and direction relative to east.
Æ
Æ
15. Find the lengths of CD and AB.
16. Find the measure of ™BCA and
Æ
the length of DE.
B
C
10
C
40
40
A
D
D
35
A
B
E
„
„
„
Let u
= ⟨0, º5⟩, v
= ⟨º2, º3⟩, and w
= ⟨4, 6⟩. Find the given sum.
17. „
u +„
v
„
18. „
u +w
„
19. „
v +w
Chapter Test
585
Page 1 of 2
Chapter Standardized Test
CHAPTER
9
TEST-TAKING STRATEGY When checking your work, try to use a method other than the one
you originally used to get your answer. If you use the same method, you may make the same
mistake twice.
1. MULTIPLE CHOICE Use the diagram to find the
6. MULTIPLE CHOICE The length of a diagonal of a
value of x.
A
¡
C
¡
E
¡
square is 16 inches. What is its perimeter?
B
¡
D
¡
6
9
A
¡
D
¡
x1
7
10
5
20
11
A
¡
B
¡
C
¡
D
¡
E
¡
11 in.
12 in.
R
2. MULTIPLE CHOICE Find the area of ⁄PQRS.
A
¡
D
¡
2
132 in.
156 in.2
B
¡
E
¡
2
143 in.
C
¡
2
154 in.
50.4 in.
B
¡
E
¡
46.4 in.
C
¡
50 in.
A
¡
D
¡
52.4 in.
4. MULTIPLE CHOICE Let the numbers represent
IV. 9, 25, 27
¡
C
¡
E
¡
¡
D
¡
A
I, II, and III only
III only
B
IV only
none
represent the side lengths of an obtuse triangle?
¡
C
¡
E
¡
A
586
A
¡
B
¡
C
¡
D
¡
E
¡
I and III only
5. MULTIPLE CHOICE Which set of numbers can
71, 70, 68
41, 39, 2
¡
D
¡
B
x
x = 8.7, y = 18.8
x = 18.8, y = 20.1
8
y
67
x = 14.4, y = 19.2
x = 12.6, y = 18.5
x = 18.8, y = 20.5
53.1°
33.7°
B
¡
E
¡
C
¡
36.9°
38.7°
56.3°
what is the measure of ™A to the nearest tenth
of a degree?
II. 7, 26, 53
0
III. 18, 21
9, 20
482 in.
9. MULTIPLE CHOICE In the diagram below,
the lengths of the sides of a triangle. Which
of the triangles are right triangles?
I. 11, 14, 3
17
302 in.
triangle is 18 centimeters long. The altitude
to the base is 12 centimeters long. What is
the approximate measure of a base angle of
the triangle?
⁄PQRS rounded to the nearest tenth.
44.4 in.
C
¡
8. MULTIPLE CHOICE The base of an isosceles
166 in.2
3. MULTIPLE CHOICE Find the perimeter of
A
¡
D
¡
162 in.
find the values of x and y. The values are rounded
to the nearest tenth.
q
S 2 in.
322 in.
B
¡
E
¡
7. MULTIPLE CHOICE Use the diagram below to
In Questions 2 and 3, use the diagram below.
P
82 in.
31.6°
B
38.0°
8
38.7°
13
C
A
51.3°
52.0°
10. MULTIPLE CHOICE Let „
v = ⟨º2, y⟩ and
30, 40, 50
25, 25, 40
17, 17, 172
Chapter 9 Right Triangles and Trigonometry
w = ⟨x, 4⟩. If „
v+„
w = ⟨6, 11⟩, what are the values
of x and y?
„
A
¡
C
¡
E
¡
x = 8, y = 8
x = 7, y = 8
x = 4, y = 7
B
¡
D
¡
x = 8, y = 7
x = º8, y = 7
Page 2 of 2
11. MULTIPLE CHOICE Points A(º8, 3) and B(1, º9) are the initial and the
Æ„
Æ„
terminal points of AB . Find the magnitude of AB .
A
¡
⟨9, º12⟩
B
¡
⟨º9, 12⟩
C
¡
D
¡
225
E
¡
15
MULTI-STEP PROBLEM In Exercises 12–14, use the diagram at the right.
⟨º7, º6⟩
25
12. Find the perimeter of the right triangle.
12
13. Find the measures of the acute angles of the right triangle.
25
14. Find the area of the shaded region.
MULTI-STEP PROBLEM In Exercises 15–18, use the diagram at the right.
Round decimals to the nearest tenth.
B
G
15. Find each of the following segment lengths: BD, BC, FG, GC, DC, and AF.
16. Find m™ABC, m™FEA, and m™BGF.
Æ
12
24
60
A
Æ
17. Find the approximate lengths of FE and AE.
103
45
D
F
C
18. Find the area of ¤ABC.
E
MULTI-STEP PROBLEM In Exercises 19–22, use the
information and the diagram.
You can use a device like the one shown to measure
the sun’s angle of elevation, ™A, above the horizon.
The variable b represents the length of the shadow
cast by a vertical marker on a ruler.
19. Find the shadow length b for each of the angle
vertical marker
5 cm
centimeter ruler
measures below.
b
a. m™A = 30°
A
angle of elevation
b. m™A = 40°
c. m™A = 50°
d. m™A = 60°
e. m™A = 70°
20. Based on your answers to Exercise 19, what happens to the value of
b as the sun rises in the sky?
21. At a certain hour, the shadow length b is 5.25 centimeters. Estimate the
sun’s angle of elevation.
22. The amount of Earth’s atmosphere that sunlight passes through depends
on the position of the sun in the sky. This amount is measured in “air
masses.” When the sun is not directly overhead, the number of air
1
sin A
masses its rays pass through is approximately . What happens to
the value of this expression as the sun approaches the horizon?
Chapter Standardized Test
587
Page 1 of 2
Cumulative Practice
CHAPTER
9
for Chapters 1–9
1. If three points all lie in two different planes, can the points be the vertices of
a triangle? Explain why or why not. (2.1)
In Exercises 2–4, use always, sometimes, or never to complete the statement.
2. Lines m, n, and t are three different coplanar lines. If m fi t and n fi t, then
? parallel. (3.1, 3.5)
line m and line n are ? represent the lengths of the sides of a
3. The numbers 8, 14, and 23 can triangle. (5.5)
? a parallelogram. (6.6)
4. A rhombus is 5.
PROOF Prove that the median to the base of an
isosceles triangle bisects the vertex angle. (4.6, 5.3)
Æ
D
B
Æ
GIVEN In ¤ABC, AB £ CB ;
Æ
A
Æ
BD is a median to AC.
C
Æ
PROVE BD bisects ™ABC.
6. In quadrilateral ABCD, m™A = 37°, m™B = 143°, and m™C = 37°.
Prove that quadrilateral ABCD is a parallelogram. (6.1, 6.3)
7. Does the design at the right have rotational symmetry? If so, describe the
rotations that map the image onto itself. (7.3)
Find the value of each variable. (3.2, 3.3, 6.5)
8.
9.
10.
y
(3x 5)
26
(x 10)
4y 25
35
(5x 7)
In Exercises 11–15, use ¤ABC with vertices A(1, º2), B(º3, º5), and C(º5, 6).
Æ
11. Find an equation of the perpendicular bisector of AC. (5.1)
12. Classify ¤ABC by its sides and by its angles. (4.1, 9.3)
13. Find the coordinates of the vertices of the image of ¤ABC after a reflection
in the y-axis. (7.2)
14. Find the coordinates of the vertices of the image of ¤ABC after a rotation of
90° counterclockwise about the origin. (7.3)
15. Find the coordinates of the image of ¤ABC after the translation
(x, y) ˘ (x º 4, y º 4) and then a reflection in the x-axis. (7.5)
Solve the proportion. (8.1)
12
5
16. = x
2
588
Chapter 9
x
3
17. = 8
7
y z
x
y
7
18. = y+3
9
Page 2 of 2
19. Rectangle ABCD has coordinates A(0, 4), B(8, 4), C(8, º2), and D(0, º2).
Æ
Æ
Points P and Q are the midpoints of AB and DC, respectively. Is rectangle
ABCD similar to rectangle APQD? Explain your answer. (8.3)
20. Show that the midsegment of a triangle forms one side of a triangle that is
similar to the original triangle. (5.4, 8.4)
21. Is a triangle with sides of lengths 6, 8, and 12 similar to a triangle with sides
of lengths 9, 12, and 18? Explain your answer. (8.5)
22. The side lengths of a triangle are 7 centimeters, 8 centimeters, and 9
centimeters. A ray bisects the largest angle, dividing the side opposite the
angle into two segments. Find the lengths of these segments. (8.6)
23. A segment has endpoints at A(6, º4) and B(12, 9). Find the images of the
1
segment under two dilations, one with scale factor and center (0, 0), and
3
1
the other with scale factor and center (0, 0). Then tell whether the two
2
image segments are parallel. (8.7)
In Exercises 24–26, use the diagram at the right.
Z
24. If XP = 2 and PY = 6, find ZP and XZ. (9.1)
25. Find XY if ZY = 23
and PY = 3. (9.1)
26. Find XY if XZ = 10 and ZY = 24. (9.2)
X
P
Y
27. Is a triangle with side lengths 12, 15, and 19 acute, right, or obtuse? (9.3)
28. A square and an equilateral triangle have sides that are 8 feet long. Find the
ratio of the length of a diagonal of the square to the height of the triangle. (9.4)
29. The lengths of the legs of a right triangle are 8 and 15. Find the sine, the
cosine, and the tangent of the smaller acute angle of the triangle. Express
each value as a fraction. (9.5)
Æ
30. In right ¤RST, m™R = 57°. The length of the hypotenuse, RS, is
20 inches. Solve the right triangle. Round decimals to the nearest tenth. (9.6)
31. Add the vectors „
u = 5, 7 and „
v = º3, 9. Draw the sum vector „
u +„
v in a
coordinate plane. Find its magnitude and its direction relative to east. (9.7)
32.
JEWELRY DESIGN A jewelry designer has a triangular piece of flat
silver. The designer wants to cut out the largest possible circle from the piece
of silver. Explain how to find the center and the radius of that circle. (5.2)
33.
CAR TRIP On the first four days of an automobile trip, a family drove
376 miles on 16 gallons of gasoline. At that rate, how many gallons of gas
will they need for the next 470 miles of their trip? (8.1)
34.
PHOTOGRAPHY A 3 inch by 5 inch photograph is enlarged so its
perimeter is 36 inches. What are the dimensions of the enlargement? (8.3)
35.
AIRPLANES Two jets leave an airport at the same time. One
jet flies west, averaging 510 miles per hour. The other jet flies
north, averaging 560 miles per hour. To the nearest tenth of a mile,
how far apart are the jets after 15 minutes? (9.2)
Cumulative Practice
589
Page 1 of 2
PROJECT
for Chapters 8 and 9
Investigating Fractals
OBJECTIVE Learn how fractals are formed and explore the properties
of fractals.
Materials: ruler, protractor, graphing calculator, graph paper
A fractal is a mathematical object that has the property of self-similarity: a part
of the shape resembles the whole shape.
Many fractals are formed by a repetitive process called iteration. The stages that
lead to a simple fractal called a Koch curve are shown below.
STAGES OF A KOCH CURVE
1
1
3
Stage 0
For stage 0, start with a
segment of length 1.
1
9
Stage 1
For stage 1, replace the
middle third of the segment
in stage 0 with two congruent
1
segments of length .
3
Stage 2
For stage 2, replace the
middle third of every
segment of stage 1 with
two congruent segments
1
of length .
9
For later stages, replace the middle third of every segment of the previous stage
with two congruent segments of the appropriate length. The fractal is the shape
that would result if you could continue this process infinitely many times.
INVESTIGATION
Copy the table below. Then answer the exercises.
Stage, n
0
1
2
3
4
Length, L
1
?
?
?
?
1. Use the diagrams above to fill in the lengths for stages 1 and 2.
2. Look for a pattern in the lengths at stages 0, 1, and 2. Use the pattern to
predict the lengths of the Koch curve at stages 3 and 4.
3. Predict what happens to the length if you repeat this iteration many times.
Explain your reasoning.
4. The length L of the fractal can be written as a function of stage n using
4 n
this formula: L = . Use a graphing calculator to sketch a graph of this
3
function. Does the graph support your answer to Exercise 3? Explain.
590
Chapter 9
Page 2 of 2
STAGES OF A KOCH SNOWFLAKE
If you join three Koch curves, you can create a
closed shape known as a Koch snowflake.
5. Sketch stage 0, stage 1, and stage 2 of the Koch
snowflake.
6. Make a table like the one on the previous page
to find the perimeter of the Koch snowflake at
stage 3 and stage 4.
7. Write a formula for the perimeter P of the Koch
snowflake at stage n.
PRESENT YOUR RESULTS
ALS
FRACT
Write a report or make a poster of your results.
• Include your drawing of the stages of the Koch
snowflake fractal.
• Include the tables you made and the answers to
Exercises 1–7.
• Describe what you learned about fractals.
EXTENSIONS
•
Research the Sierpinski Triangle or the Cantor Set. Sketch the first four
stages of these fractals.
•
Create a fractal of your own. Start with a geometric shape and
perform some alteration or transformation. Show the first three stages
of your fractal.
•
Suppose a Koch snowflake has
an area of 1 at stage 0. Find the
area of the snowflake at stage 1
and stage 2.
•
Research the Mandelbrot Set and
find some images of it. Many
Web sites on the Internet feature
images of this fractal.
The Mandelbrot Set reveals
dramatic images as you
zoom in to see details.
Project
591