6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 434
CHAPTER 5 Analytic Trigonometry
434
5.5 The Law of Sines
What you’ll learn about
• Deriving the Law of Sines
• Solving Triangles (AAS, ASA)
• The Ambiguous Case (SSA)
• Applications
... and why
The Law of Sines is a powerful
extension of the triangle
congruence theorems of
Euclidean geometry.
Deriving the Law of Sines
Recall from geometry that a triangle has six parts (three sides (S), three angles (A)),
but that its size and shape can be completely determined by fixing only three of those
parts, provided they are the right three. These threesomes that determine triangle congruence are known by their acronyms: AAS, ASA, SAS, and SSS. The other two
acronyms represent matchups that don’t quite work: AAA determines similarity only,
while SSA does not even determine similarity.
With trigonometry we can find the other parts of the triangle once congruence is established. The tools we need are the Law of Sines and the Law of Cosines, the subjects of
our last two trigonometric sections.
The Law of Sines states that the ratio of the sine of an angle to the length of its opposite side is the same for all three angles of any triangle.
Law of Sines
In any ¢ABC with angles A, B, and C opposite sides a, b, and c, respectively,
the following equation is true:
sin A
sin B
sin C
=
=
.
a
c
b
The derivation of the Law of Sines refers to the two triangles in Figure 5.13, in each of
which we have drawn an altitude to side c. Right triangle trigonometry applied to either
of the triangles in Figure 5.13 tells us that
sin A =
h
.
b
sin B =
h
,
a
In the acute triangle on the top,
while in the obtuse triangle on the bottom,
C
b
A
c
B
C
b
A
c
a
sin 1p - B2 =
a
h
h
h
.
a
But sin 1p - B2 = sin B, so in either case
h
.
a
Solving for h in both equations yields h = b sin A = a sin B. The equation
b sin A = a sin B is equivalent to
sin B =
sin B
sin A
=
.
a
b
B
FIGURE 5.13 The Law of Sines.
If we were to draw an altitude to side a and repeat the same steps as above, we would
reach the conclusion that
sin C
sin B
=
.
c
b
Putting the results together,
sin C
sin B
sin A
=
=
.
a
c
b
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 435
SECTION 5.5
The Law of Sines
435
Solving Triangles (AAS, ASA)
Two angles and a side of a triangle, in any order, determine the size and shape of a triangle completely. Of course, two angles of a triangle determine the third, so we really get
one of the missing three parts for free. We solve for the remaining two parts (the unknown sides) with the Law of Sines.
EXAMPLE 1 Solving a Triangle Given Two Angles and a Side
Solve ¢ABC given that ∠A = 36°, ∠B = 48°, and a = 8. (See Figure 5.14.)
SOLUTION First, we note that ∠C = 180° - 36° - 48° = 96°.
C
b
8
36°
48°
A
c
We then apply the Law of Sines:
B
FIGURE 5.14 A triangle determined by
AAS. (Example 1)
sin A
sin B
and
=
a
b
sin 36°
sin 48°
=
8
b
8 sin 48°
b =
sin 36°
b L 10.115
The six parts of the triangle are:
∠A = 36°
∠B = 48°
∠C = 96°
sin A
sin C
=
a
c
sin 36°
sin 96°
=
c
8
8 sin 96°
c =
sin 36°
c L 13.536
a = 8
b L 10.115
c L 13.536
Now try Exercise 1.
The Ambiguous Case (SSA)
While two angles and a side of a triangle are always sufficient to determine its size and
shape, the same cannot be said for two sides and an angle. Perhaps unexpectedly, it depends on where that angle is. If the angle is included between the two sides (the SAS case),
then the triangle is uniquely determined up to congruence. If the angle is opposite one of
the sides (the SSA case), then there might be one, two, or zero triangles determined.
A
B
FIGURE 5.15 The diagram for
part 1. (Exploration 1)
EXPLORATION 1
B
h
A
Solving a triangle in the SAS case involves the Law of Cosines and will be handled in
the next section. Solving a triangle in the SSA case is done with the Law of Sines, but
with an eye toward the possibilities, as seen in the following Exploration.
Determining the Number of Triangles
We wish to construct ¢ABC given angle A, side AB, and side BC.
1. Suppose ∠A is obtuse and that side AB is as shown in Figure 5.15. To complete
FIGURE 5.16 The diagram for
part 2. (Exploration 1)
2. Suppose ∠A is acute and that side AB is as shown in Figure 5.16. To complete
B
h
A
the triangle, side BC must determine a point on the dotted horizontal line (which
extends infinitely to the left). Explain from the picture why a unique triangle
¢ABC is determined if BC 7 AB, but no triangle is determined if BC … AB.
C
FIGURE 5.17 The diagram for
parts 3–5. (Exploration 1)
the triangle, side BC must determine a point on the dotted horizontal line (which
extends infinitely to the right). Explain from the picture why a unique triangle
¢ABC is determined if BC = h, but no triangle is determined if BC 6 h.
3. Suppose ∠A is acute and that side AB is as shown in Figure 5.17. If
AB 7 BC 7 h, then we can form a triangle as shown. Find a second point C
on the dotted horizontal line that gives a side BC of the same length, but determines a different triangle. (This is the “ambiguous case.”)
4. Explain why sin C is the same in both triangles in the ambiguous case. (This is
why the Law of Sines is also ambiguous in this case.)
5. Explain from Figure 5.17 why a unique triangle is determined if BC Ú AB.
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 436
CHAPTER 5 Analytic Trigonometry
436
Now that we know what can happen, let us try the algebra.
EXAMPLE 2 Solving a Triangle Given Two Sides and an Angle
Solve ¢ABC given that a = 7, b = 6, and ∠ A = 26.3°. (See Figure 5.18.)
C
7
6
SOLUTION By drawing a reasonable sketch (Figure 5.18), we can assure our-
26.3°
A
c
B
FIGURE 5.18 A triangle determined
by SSA. (Example 2)
selves that this is not the ambiguous case. (In fact, this is the case described in step 5
of Exploration 1.)
Begin by solving for the acute angle B, using the Law of Sines:
sin A
sin B
=
.
a
b
sin 26.3°
sin B
=
7
6
6 sin 26.3°
sin B =
7
6 sin 26.3°
B = sin-1 a
b
7
B = 22.3°
Law of Sines
Round to match accuracy of given angle.
Then, find the obtuse angle C by subtraction:
C = 180° - 26.3° - 22.3°
= 131.4°
Finally, find side c:
sin A
sin C
=
a
c
sin 26.3°
sin 131.4°
=
c
7
7 sin 131.4°
c =
sin 26.3°
c L 11.9
The six parts of the triangle are:
∠A = 26.3°
a = 7
∠ B = 22.3°
b = 6
∠ C = 131.4°
c L 11.9
Now try Exercise 9.
EXAMPLE 3 Handling the Ambiguous Case
Solve ¢ABC given that a = 6, b = 7, and ∠ A = 30°.
C
30°
7
SOLUTION By drawing a reasonable sketch (Figure 5.19), we see that two trian-
6
A
c
gles are possible with the given information. We keep this in mind as we proceed.
B
(a)
C
30°
7
6
A c B
(b)
FIGURE 5.19 Two triangles determined
by the same SSA values. (Example 3)
We begin by using the Law of Sines to find angle B.
sin A
sin B
=
a
b
sin 30°
sin B
=
6
7
7 sin 30°
sin B =
6
7 sin 30°
B = sin-1 a
b
6
B = 35.7°
Law of Sines
Round to match accuracy of given angle.
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 437
SECTION 5.5
The Law of Sines
437
Notice that the calculator gave us one value for B, not two. That is because we used
the function sin-1, which cannot give two output values for the same input value.
Indeed, the function sin-1 will never give an obtuse angle, which is why we chose
to start with the acute angle in Example 2. In this case, the calculator has found the
angle B shown in Figure 5.19a.
Find the obtuse angle C by subtraction:
C = 180° - 30.0° - 35.7° = 114.3°
Finally, find side c:
sin A
sin C
=
a
c
sin 30.0°
sin 114.3°
=
c
6
c =
6 sin 114.3°
sin 30°
c L 10.9
So, under the assumption that angle B is acute (see Figure 5.19a), the six parts of the
triangle are:
∠A = 30.0°
∠B = 35.7°
a = 6
b = 7
∠C = 114.3°
c L 10.9
If angle B is obtuse, then we can see from Figure 5.19b that it has measure
180° - 35.7° = 144.3°.
By subtraction, the acute angle C = 180° - 30.0° - 144.3° = 5.7°. We then recompute c:
c =
6 sin 5.7°
L 1.2
sin 30°
Substitute 5.7° for 114.3° in earlier computation.
So, under the assumption that angle B is obtuse (see Figure 5.19b), the six parts of
the triangle are:
∠A = 30.0°
a = 6
∠B = 144.3°
b = 7
∠C = 5.7°
c L 1.2
Now try Exercise 19.
Applications
Many problems involving angles and distances can be solved by superimposing a triangle onto the situation and solving the triangle.
N
N
EXAMPLE 4 Locating a Fire
C
b
32°
A
h
a
Forest Ranger Chris Johnson at ranger station A sights a fire in the direction 32° east
of north. Ranger Rick Thorpe at ranger station B, 10 miles due east of A, sights the
same fire on a line 48° west of north. Find the distance from each ranger station to
the fire.
48°
10 mi
B
FIGURE 5.20 Determining the
location of a fire. (Example 4)
SOLUTION Let C represent the location of the fire. A sketch (Figure 5.20) shows
the superimposed triangle, ¢ ABC, in which angles A and B and their included side
(AB) are known. This is a setup for the Law of Sines.
(continued)
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 438
CHAPTER 5 Analytic Trigonometry
438
Note that ∠A = 90° - 32° = 58° and ∠B = 90° - 48° = 42°. By subtraction,
we find that ∠C = 180° - 58° - 42° = 80°.
sin B
sin C
sin A
sin C
=
=
and
Law of Sines
a
c
c
b
sin 58°
sin 80°
sin 42°
sin 80°
=
=
a
10
b
10
10 sin 58°
10 sin 42°
a =
b =
sin 80°
sin 80°
a L 8.6
b L 6.8
The fire is about 6.8 miles from ranger station A and about 8.6 miles from ranger station B.
Now try Exercise 45.
EXAMPLE 5 Finding the Height of a Pole
b
A road slopes 10° above the horizontal, and a vertical telephone pole stands beside
the road. The angle of elevation of the Sun is 62°, and the pole casts a 14.5-foot
shadow downhill along the road. Find the height of the telephone pole.
6 2°
C
a
SOLUTION This is an interesting variation on a typical application of right triangle
trigonometry. The slope of the road eliminates the convenient right angle, but we can
still solve the problem by solving a triangle.
1 0°
A
c
B
Figure 5.21 shows the superimposed triangle, ¢ABC. A little preliminary geometry
is required to find the measure of angles A and C. Due to the slope of the road, angle
A is 10° less than the angle of elevation of the Sun and angle B is 10° more than a
right angle. That is,
FIGURE 5.21 A telephone pole on
a slope. (Example 5)
∠A = 62° - 10° = 52°
∠B = 90° + 10° = 100°
∠C = 180° - 52° - 100° = 28°
Therefore,
sin C
sin A
=
a
c
sin 28°
sin 52°
=
a
14.5
14.5 sin 52°
a =
sin 28°
a L 24.3
Law of Sines
Round to match accuracy of input.
Now try Exercise 39.
The pole is approximately 24.3 feet high.
QUICK REVIEW 5.5
(For help, go to Sections 4.2 and 4.7.)
Exercise numbers with a gray background indicate problems
that the authors have designed to be solved without a calculator.
In Exercises 1–4, solve the equation a/b = c/d for the given variable.
1. a
2. b
3. c
4. d
In Exercises 5 and 6, evaluate the expression.
5.
7 sin 48°
sin 23°
6.
9 sin 121°
sin 14°
In Exercises 7–10, solve for the angle x.
7. sin x = 0.3,
0° 6 x 6 90°
8. sin x = 0.3,
90° 6 x 6 180°
9. sin x = - 0.7,
180° 6 x 6 270°
10. sin x = - 0.7,
270° 6 x 6 360°
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 439
SECTION 5.5
The Law of Sines
439
SECTION 5.5 EXERCISES
25.
In Exercises 1–4, solve the triangle.
1.
2.
C
a
3.7
60°
A
B
45°
c
120°
C
b
B
3.
22
B
a
19
56°
A
23
(a)
a
19
56°
A
b
(b)
C
26. B
4.
B
C
A
17
15°
a
B
a
35° 100°
b
A
81
C
119° 29°
b
119°
A
A
In Exercises 5–8, solve the triangle.
A
(b)
In Exercises 27–36, respond in one of the following ways:
(a) State, “Cannot be solved with the Law of Sines.”
5. A = 40°,
B = 30°,
b = 10
(b) State, “No triangle is formed.”
6. A = 50°,
B = 62°,
a = 4
(c) Solve the triangle.
7. A = 33°,
B = 70°, b = 7
27. A = 61°, a = 8,
b = 21
8. B = 16°,
C = 103°,
28. B = 47°,
b = 21
c = 12
29. A = 136°,
In Exercises 9–12, solve the triangle.
a = 8,
a = 15,
b = 28
9. A = 32°,
a = 17, b = 11
30. C = 115°,
10. A = 49°,
a = 32, b = 28
31. B = 42°,
c = 18,
C = 39°
11. B = 70°,
b = 14,
32. A = 19°,
b = 22,
B = 47°
33. C = 75°,
b = 49,
c = 48
12. C = 103°,
89
C
(a)
81°
c
c
81
92
40°
B
B
c
C
c
C
c = 9
b = 46, c = 61
In Exercises 13–18, state whether the given measurements determine
zero, one, or two triangles.
b = 12,
34. A = 54°, a = 13,
35. B = 31°, a = 8,
c = 7
b = 15
c = 11
13. A = 36°,
a = 2, b = 7
36. C = 65°, a = 19,
14. B = 82°,
b = 17,
c = 15
15. C = 36°,
a = 17,
c = 16
16. A = 73°,
a = 24, b = 28
37. Surveying a Canyon Two markers A and B on the
same side of a canyon rim are 56 ft apart. A third marker C,
located across the rim, is positioned so that ∠ BAC = 72°
and ∠ABC = 53°.
17. C = 30°,
a = 18, c = 9
18. B = 88°,
b = 14,
A
c = 62
In Exercises 19–22, two triangles can be formed using the given measurements. Solve both triangles.
19. A = 64°,
a = 16,
b = 17
20. B = 38°,
b = 21,
c = 25
21. C = 68°,
a = 19,
c = 18
22. B = 57°,
a = 11,
b = 10
(b) One triangle
(b) One triangle
B
(b) Find the distance between the two canyon rims. (Assume
they are parallel.)
(c) Zero triangles
24. Determine the values of c that will produce the given number
of triangles if b = 12 and C = 53°.
(a) Two triangles
56 ft
C
(a) Find the distance between C and A.
23. Determine the values of b that will produce the given number
of triangles if a = 10 and B = 42°.
(a) Two triangles
b = 22
(c) Zero triangles
In Exercises 25 and 26, decide whether the triangle can be solved using
the Law of Sines. If so, solve it. If not, explain why not.
38. Weather Forecasting
N
N
Two meteorologists are 25 mi
C
apart located on an east-west road.
The meteorologist at point A sights
b
a
a tornado 38° east of north. The
38°
53°
h
meteorologist at point B sights the
A
B
same tornado 53° west of north.
25 mi
Find the distance from each meteorologist to the tornado. Also find the distance between the
tornado and the road.
6965_CH05_pp403-454.qxd
440
1/14/10
1:52 PM
Page 440
CHAPTER 5 Analytic Trigonometry
39. Engineering Design A
vertical flagpole stands beside
a road that slopes at an angle
of 15° with the horizontal.
When the angle of elevation of
the Sun is 62°, the flagpole
casts a 16-ft shadow downhill
along the road. Find the height
of the flagpole.
A
62°
C
33°
62°
b
75°
105°
A
20 mi
28° a
16 ft
B
52°
15°
Horizontal
40. Altitude Observers 2.32 mi
apart see a hot-air balloon
directly between them but at the
angles of elevation shown in the
figure. Find the altitude of the
balloon.
B
46. Using Measurement Data A geometry class is divided
into ten teams, each of which is given a yardstick and a protractor to find the distance from a point A on the edge of a pond to a
tree at a point C on the opposite shore. After they mark points A
and B with stakes, each team uses a protractor to measure angles
A and B and a yardstick to measure distance AB. Their measurements are given in the table.
28°
A
A
37°
2.32 mi
B
41. Reducing Air Resistance A 4-ft airfoil attached to the
cab of a truck reduces wind resistance. If the angle between the
airfoil and the cab top is 18° and angle B is 10°, find the length
of a vertical brace positioned as shown in the figure.
B
79°
81°
79°
80°
79°
18°
B
AB
26 ¿
25 ¿
26 ¿
26 ¿
25 ¿
84°
82°
83°
87°
87°
A
4 ft
S (ship)
B
83°
82°
78°
77°
79°
4–
5–
0–
1–
11–
AB
25 ¿
26 ¿
25 ¿
26 ¿
25 ¿
84°
82°
85°
83°
82°
3–
5–
8–
4–
7–
Use the data to find the class’s best estimate for the distance
AC.
C
42. Group Activity Ferris Wheel Design A Ferris
wheel has 16 evenly spaced cars. The distance between adjacent chairs is 15.5 ft. Find the radius of the wheel (to the nearest 0.1 ft).
43. Finding Height Two observers are 600 ft apart on
opposite sides of a flagpole. The angles of elevation from the
observers to the top of the pole are 19° and 21°. Find the
height of the flagpole.
44. Finding Height Two observers are 400 ft apart on
opposite sides of a tree. The angles of elevation from the
observers to the top of the tree are 15° and 20°. Find the
height of the tree.
45. Finding Distance Two lighthouses A and B are known
to be exactly 20 mi apart on a north-south line. A ship’s captain
at S measures ∠ASB to be 33°. A radio operator at B measures
∠ABS to be 52°. Find the distance from the ship to each lighthouse.
A
B
Standardized Test Questions
47. True or False The ratio of the sines of any two angles in a triangle equals the ratio of the lengths of their
opposite sides. Justify your answer.
48. True or False The perimeter of a triangle with two
10-inch sides and two 40° angles is greater than 36.
Justify your answer.
You may use a graphing calculator when answering these questions.
49. Multiple Choice The
length x in the triangle shown
at the right is
(A) 8.6.
(D) 19.2.
(B) 15.0.
(E) 22.6.
(C) 18.1.
x
95°
12.0
53°
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 441
SECTION 5.5
50. Multiple Choice Which of the following three triangle
parts do not necessarily determine the other three parts?
(A) AAS
(B) ASA
(D) SSA
(E) SSS
(C) SAS
51. Multiple Choice The shortest side of a triangle with
angles 50°, 60°, and 70° has length 9.0. What is the length of
the longest side?
(A) 11.0
(B) 11.5
(D) 12.5
(E) 13.0
(C) 12.0
(B) One
(D) Three
(E) Infinitely many
Extending the Ideas
C
(C) Two
5
22°
B
A
53. Writing to Learn
(a) Show that there are infinitely many triangles with AAA
given if the sum of the three positive angles is 180°.
(b) Give three examples of triangles where A = 30°, B = 60°,
and C = 90°.
(c) Give three examples where A = B = C = 60°.
54. Use the Law of Sines and the cofunction identities to derive the
following formulas from right triangle trigonometry:
opp
hyp
(d) In terms of the given angle A and the given length AB, state
the conditions on length BC that will result in two possible
triangles being formed.
8
Explorations
(a) sin A =
(c) In terms of the given angle A and the given length AB, state
the conditions on length BC that will result in a unique triangle being formed.
56. Solve this triangle assuming that ∠B is obtuse.
[Hint: Draw a perpendicular from A to the line
through B and C.]
52. Multiple Choice How many noncongruent triangles
ABC can be formed if AB = 5, A = 60°, and BC = 8?
(A) None
441
The Law of Sines
(b) cos A =
adj
hyp
(c) tan A =
opp
adj
55. Wrapping up Exploration 1 Refer to Figures 5.16
and 5.17 in Exploration 1 of this section.
(a) Express h in terms of angle A and length AB.
(b) In terms of the given angle A and the given length AB, state
the conditions on length BC that will result in no triangle
being formed.
57. Pilot Calculations Towers A and B are known to be
4.1 mi apart on level ground. A pilot measures the angles of
depression to the towers to be 36.5° and 25°, respectively, as
shown in the figure. Find distances AC and BC and the height
of the airplane.
C
25°
36.5°
A 4.1 mi B
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 442
CHAPTER 5 Analytic Trigonometry
442
5.6 The Law of Cosines
What you’ll learn about
• Deriving the Law of Cosines
• Solving Triangles (SAS, SSS)
• Triangle Area and Heron’s
Formula
• Applications
Deriving the Law of Cosines
Having seen the Law of Sines, you will probably not be surprised to learn that there is
a Law of Cosines. There are many such parallels in mathematics. What you might
find surprising is that the Law of Cosines has absolutely no resemblance to the Law
of Sines. Instead, it resembles the Pythagorean Theorem. In fact, the Law of Cosines
is often called the “generalized Pythagorean Theorem” because it contains that classic
theorem as a special case.
... and why
The Law of Cosines is an
important extension of the
Pythagorean Theorem, with
many applications.
Law of Cosines
Let ¢ABC be any triangle with sides and angles labeled in the usual way
(Figure 5.22).
Then
a 2 = b 2 + c2 - 2bc cos A
b 2 = a 2 + c2 - 2ac cos B
c2 = a 2 + b 2 - 2ab cos C
We derive only the first of the three equations, since the other two are derived in
exactly the same way. Set the triangle in a coordinate plane so that the angle that
appears in the formula (in this case, A) is at the origin in standard position, with side
c along the positive x-axis. Depending on whether angle A is right (Figure 5.23a),
acute (Figure 5.23b), or obtuse (Figure 5.23c), the point C will be on the y-axis, in
QI, or in QII.
a
A
c
a
b
B
FIGURE 5.22 A triangle with the usual
labeling (angles A, B, C; opposite
sides a, b, c).
A
C(x, y)
C(x, y)
C(x, y)
C
b
y
y
y
c
b
B(c, 0)
x
A
a
c
B(c, 0)
x
A
c
B(c, 0)
x
(c)
(b)
(a)
a
b
FIGURE 5.23 Three cases for proving the Law of Cosines.
In each of these three cases, C is a point on the terminal side of angle A in standard
position, at distance b from the origin. Denote the coordinates of C by (x, y). By
our definitions for trigonometric functions of any angle (Section 4.3), we can
conclude that
x
= cos A
b
and
y
= sin A,
b
x = b cos A
and
y = b sin A.
and therefore
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 443
SECTION 5.6
The Law of Cosines
443
Now set a equal to the distance from C to B using the distance formula:
a = 21x - c22 + 1y - 022
Distance formula
a 2 = 1x - c22 + y 2
Square both sides.
= 1b cos A - c22 + 1b sin A22
Substitution
= b 2 cos2 A - 2bc cos A + c2 + b 2 sin2 A
= b 21cos2 A + sin2 A2 + c2 - 2bc cos A
= b 2 + c2 - 2bc cos A
Pythagorean identity
Solving Triangles (SAS, SSS)
While the Law of Sines is the tool we use to solve triangles in the AAS and ASA cases,
the Law of Cosines is the required tool for SAS and SSS. (Both methods can be used in
the SSA case, but remember that there might be 0, 1, or 2 triangles.)
EXAMPLE 1 Solving a Triangle (SAS)
C
Solve ¢ABC given that a = 11, b = 5, and C = 20°. (See Figure 5.24.)
5
SOLUTION
20°
A
c2 = a 2 + b 2 - 2ab cos C
11
= 112 + 52 - 21112152 cos 20°
c
= 42.6338 Á
c = 142.6338 Á L 6.5
B
FIGURE 5.24 A triangle with two sides and
an included angle known. (Example 1)
We could now use either the Law of Cosines or the Law of Sines to find one of the
two unknown angles. As a general rule, it is better to use the Law of Cosines to find
angles, since the arccosine function will distinguish obtuse angles from acute angles.
a 2 = b 2 + c2 - 2bc cos A
112 = 52 + 16.529 Á 22 - 215216.529 Á 2cos A
cos A =
52 + 16.529 Á 22 - 112
215216.529 Á 2
A = cos -1 a
L 144.8°
52 + 16.529 Á 22 - 112
215216.529 Á 2
b
B = 180° - 144.8° - 20°
= 15.2°
So the six parts of the triangle are:
A = 144.8°
B = 15.2°
C = 20°
A
5
7
C
9
B
FIGURE 5.25 A triangle with three
sides known. (Example 2)
a = 11
b = 5
c L 6.5
EXAMPLE 2 Solving a Triangle (SSS)
Solve ¢ABC if a = 9, b = 7, and c = 5. (See Figure 5.25.)
Now try Exercise 1.
6965_CH05_pp403-454.qxd
2/2/10
11:26 AM
Page 444
CHAPTER 5 Analytic Trigonometry
444
SOLUTION We use the Law of Cosines to find two of the angles. The third angle
can be found by subtraction from 180°.
a 2 = b 2 + c2 - 2bc cos A
b 2 = a 2 + c2 - 2ac cos B
9 2 = 72 + 52 - 2172152cos A
72 = 9 2 + 52 - 2192152cos B
70 cos A = - 7
90 cos B = 57
A = cos - 1 1-0.12
B = cos - 1 157/902
L 95.7°
L 50.7°
Then C = 180° - 95.7° - 50.7° = 33.6°.
Now try Exercise 3.
Triangle Area and Heron’s Formula
The same parts that determine a triangle also determine its area. If the parts happen to
be two sides and an included angle (SAS), we get a simple area formula in terms of
those three parts that does not require finding an altitude.
Observe in Figure 5.23 (used in explaining the Law of Cosines) that each triangle has
base c and altitude y = b sin A. Applying the standard area formula, we have
¢ Area =
1
1
1
1base21height2 = 1c21b sin A2 = bc sin A.
2
2
2
This is actually three formulas in one, as it does not matter which side we use as the
base.
Area of a Triangle
¢ Area =
1
1
1
bc sin A = ac sin B = ab sin C
2
2
2
EXAMPLE 3 Finding the Area of a Regular Polygon
9
θ
Find the area of a regular octagon (8 equal sides, 8 equal angles) inscribed inside a
circle of radius 9 inches.
9
SOLUTION Figure 5.26 shows that we can split the octagon into 8 congruent trian-
gles. Each triangle has two 9-inch sides with an included angle of u = 360°/8 = 45°.
The area of each triangle is
FIGURE 5.26 A regular octagon
inscribed inside a circle of radius 9 inches.
(Example 3)
Heron’s Formula
The formula is named after Heron of Alexandria,
whose proof of the formula is the oldest on
record, but ancient Arabic scholars claimed to
have known it from the works of Archimedes
of Syracuse centuries earlier. Archimedes
(c. 287–212 B.C.E.) is considered to be the greatest mathematician of all antiquity.
¢ Area = 11/22192192 sin 45° = 181/2) sin 45° = 8112/4.
Therefore, the area of the octagon is
Area = 8 ¢ Area = 16212 L 229 square inches.
Now try Exercise 31.
There is also an area formula that can be used when the three sides of the triangle are
known.
Although Heron proved this theorem using only classical geometric methods, we prove
it, as most people do today, by using the tools of trigonometry.
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 445
SECTION 5.6
The Law of Cosines
445
THEOREM Heron’s Formula
Let a, b, and c be the sides of ¢ABC, and let s denote the semiperimeter.
1a + b + c)/2.
Then the area of ¢ABC is given by Area = 2s1s - a21s - b21s - c2.
Proof
1
ab sin C
2
41Area2 = 2ab sin C
Area =
161Area)2 = 4a 2b 2 sin2 C
= 4a 2b 211 - cos2 C2
2 2
2 2
Pythagorean identity
2
= 4a b - 4a b cos C
= 4a 2b 2 - 12ab cos C22
= 4a 2b 2 - 1a 2 + b 2 - c222
2
2
2
Law of Cosines
2
2
2
= 12ab - 1a + b - c 2212ab + 1a + b - c 22
Difference of squares
= 1c2 - 1a 2 - 2ab + b 22211a 2 + 2ab + b 22 - c22
= 1c2 - (a - b22211a + b22 - c22
= 1c - 1a - b221c + (a - b2211a + b2 - c211a + b2 + c2
Difference of squares
= 1c - a + b21c + a - b21a + b - c21a + b + c2
= 12s - 2a212s - 2b212s - 2c212s2
2s = a + b + c
161Area22 = 16s1s - a21s - b21s - c2
1Area22 = s1s - a21s - b21s - c2
Area = 2s1s - a21s - b21s - c2
EXAMPLE 4 Using Heron’s Formula
Find the area of a triangle with sides 13, 15, 18.
SOLUTION First we compute the semiperimeter: s = 113 + 15 + 182/2 = 23.
Then we use Heron’s Formula
Area = 123 123 - 132123 - 152123 - 182
= 123 # 10 # 8 # 5 = 19200 - 20123.
The approximate area is 96 square units.
Now try Exercise 21.
Applications
We end this section with a few applications.
EXAMPLE 5 Measuring a Baseball Diamond
The bases on a baseball diamond are 90 feet apart, and the front edge of the pitcher’s
rubber is 60.5 feet from the back corner of home plate. Find the distance from the
center of the front edge of the pitcher’s rubber to the far corner of first base.
(continued)
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 446
CHAPTER 5 Analytic Trigonometry
446
SOLUTION Figure 5.27 shows first base as A, the pitcher’s rubber as B, and home
Second base
plate as C. The distance we seek is side c in ¢ABC.
By the Law of Cosines,
c2 = 60.52 + 90 2 - 2160.521902 cos 45°
c
B
Third
base
A (Fir
c = 260.52 + 90 2 - 2160.521902 cos 45°
L 63.7
60.5 ft
45∞
90 ft
The distance from first base to the pitcher’s rubber is about 63.7 feet.
Now try Exercise 37.
C
(Home plate)
FIGURE 5.27 The diamond-shaped part
of a baseball diamond. (Example 5)
Platonic Solids
The regular tetrahedron in Example 6 is one of
only 5 regular solids (solids with faces that are
congruent polygons having equal angles and
equal sides). The others are the cube (6 square
faces), the octahedron (8 triangular faces), the
dodecahedron (12 pentagonal faces), and the
icosahedron (20 triangular faces). Referred to as
the Platonic solids, Plato did not discover them,
but they are featured in his cosmology as being
the stuff of which everything in the universe is
made. The Platonic universe itself is a dodecahedron, a favorite symbol of the Pythagoreans.
2
2
2
3
3
D
1
B
A regular tetrahedron is a solid with four faces, each of which is an equilateral triangle. Find the measure of the dihedral angle formed along the common edge of two
intersecting faces of a regular tetrahedron with edges of length 2.
SOLUTION Figure 5.28 shows the tetrahedron. Point B is the midpoint of edge
DE, and A and C are the endpoints of the opposite edge. The measure of ∠ABC is
the same as the measure of the dihedral angle formed along edge DE, so we will find
the measure of ∠ABC.
Because both ¢ADB and ¢CDB are 30° - 60° - 90° triangles, AB and BC both
have length 13. If we apply the Law of Cosines to ¢ABC, we obtain
22 = 11322 + 11322 - 213 13 cos 1∠ ABC2
cos 1 ∠ABC2 =
1
3
1
∠ABC = cos -1 a b L 70.53°
3
C
A
EXAMPLE 6 Measuring a Dihedral Angle (Solid Geometry)
1
E
FIGURE 5.28 The measure of ∠ ABC is
the same as the measure of any dihedral
angle formed by two of the tetrahedron’s
faces. (Example 6)
The dihedral angle has the same measure as ∠ABC, approximately 70.53°. (We
chose sides of length 2 for computational convenience, but in fact this is the measure
of a dihedral angle in a regular tetrahedron of any size.)
Now try Exercise 43.
EXPLORATION 1
Estimating Acreage of a Plot of Land
Jim and Barbara are house hunting and need to estimate the size of an irregular adjacent lot that is described by the owner as “a little more than an acre.” With Barbara
stationed at a corner of the plot, Jim starts at another corner and walks a straight line
toward her, counting his paces. They then shift corners and Jim paces again, until
they have recorded the dimensions of the lot (in paces) as in Figure 5.29. They later
measure Jim’s pace as 2.2 feet. What is the approximate acreage of the lot?
1. Use Heron’s Formula to find the area in square paces.
2. Convert the area to square feet, using the measure of Jim’s pace.
3. There are 5280 feet in a mile. Convert the area to square miles.
4. There are 640 square acres in a square mile. Convert the area to acres.
5. Is there good reason to doubt the owner’s estimate of the acreage of the lot?
6. Would Jim and Barbara be able to modify their system to estimate the area of
an irregular lot with five straight sides?
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 447
SECTION 5.6
The Law of Cosines
447
102
81
112
86
115
FIGURE 5.29 Dimensions (in paces) of an irregular plot of land. (Exploration 1)
Chapter Opener Problem (from page 403)
Problem: Because deer require food, water, cover for protection from weather
and predators, and living space for healthy survival, there are natural limits to the
number of deer that a given plot of land can support. Deer populations in national
parks average 14 animals per square kilometer. If a triangular region with sides of
3 kilometers, 4 kilometers, and 6 kilometers has a population of 50 deer, how close
is the population on this land to the average national park population?
Solution: We can find the area of the land region
6 km
3 km
4 km
by using Heron’s Formula with
and
s = 13 + 4 + 62/2 = 13/2
Area = 1s1s - a21s - b21s - c2
=
=
13 13
13
13
a
- 3b a
- 4b a
- 6b
B2
2
2
2
13 7 5 1
a b a b a b L 5.3,
B2 2 2 2
so the area of the land region is 5.3 km2.
If this land were to support 14 deer/km2, it would have
15.3 Á km22114 deer/km22 = 74.7 L 75 deer. Thus, the land supports 25 deer
less than the average.
QUICK REVIEW 5.6
(For help, go to Sections 2.4 and 4.7.)
Exercise numbers with a gray background indicate problems
that the authors have designed to be solved without a calculator.
In Exercises 1– 4, find an angle between 0° and 180° that is a solution to the equation.
1. cos A = 3/5
2. cos C = - 0.23
3. cos A = - 0.68
4. 3 cos C = 1.92
In Exercises 5 and 6, solve the equation (in terms of x and y) for
(a) cos A and (b) A, 0 … A … 180°.
5. 9 2 = x 2 + y 2 - 2xy cos A 6. y 2 = x 2 + 25 - 10 cos A
In Exercises 7–10, find a quadratic polynomial with real coefficients
that satisfies the given condition.
7. Has two positive zeros
8. Has one positive and one negative zero
9. Has no real zeros
10. Has exactly one positive zero
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 448
CHAPTER 5 Analytic Trigonometry
448
SECTION 5.6 EXERCISES
31. Find the area of a regular hexagon inscribed in a circle of
radius 12 inches.
In Exercises 1–4, solve the triangle.
1.
2.
B
8
131°
A
13
C
3.
4.
A
24
B
32. Find the area of a regular nonagon (9 sides) inscribed in a
circle of radius 10 inches.
12
A
C
42°
C
19
14
B
35
A
17
28
B
27
C
In Exercises 5–16, solve the triangle.
5. A = 55°,
b = 12,
c = 7
6. B = 35°,
a = 43,
c = 19
7. a = 12,
b = 21,
8. b = 22,
C = 95°
c = 31,
A = 82°
9. a = 1,
b = 5,
c = 4
10. a = 1,
b = 5,
c = 8
11. a = 3.2,
b = 7.6,
12. a = 9.8,
b = 12,
c = 23
13. A = 42°,
a = 7,
b = 10
14. A = 57°,
a = 11,
b = 10
15. A = 63°,
a = 8.6,
b = 11.1
16. A = 71°,
a = 9.3,
b = 8.5
c = 6.4
33. Find the area of a regular hexagon circumscribed about a circle
of radius 12 inches. [Hint: Start by finding the distance from a
vertex of the hexagon to the center of the circle.]
34. Find the area of a regular nonagon (9 sides) circumscribed
about a circle of radius 10 inches.
35. Measuring Distance Indirectly
Juan wants to find the distance
between two points A and B
on opposite sides of a building.
A
He locates a point C that is 110 ft
from A and 160 ft from B, as
illustrated in the figure. If the
angle at C is 54°, find distance AB. 110 ft
36. Designing a Baseball Field
b = 32 ft,
c = 19 ft
18. A = 52°,
b = 14 m,
c = 21 m
19. B = 101°,
a = 10 cm,
c = 22 cm
20. C = 112°,
a = 1.8 in.,
b = 5.1 in.
In Exercises 21–28, decide whether a triangle can be formed with the
given side lengths. If so, use Heron’s Formula to find the area of the
triangle.
21. a = 4,
b = 5,
c = 8
22. a = 5,
b = 9,
c = 7
23. a = 3,
b = 5,
c = 8
24. a = 23,
25. a = 19.3,
26. a = 8.2,
b = 19,
b = 12.5,
54°
(b) Find ∠ B in ¢ABC.
Second base
Third
base
c
B
A (First base)
60.5 ft
45°
90 ft
C
(Home plate)
c = 12
b = 22.5,
160 ft
(a) Find the distance from the
C
center of the front edge of the
pitcher’s rubber to the far
corner of second base. How does this distance compare
with the distance from the pitcher’s rubber to first base?
(See Example 5.)
In Exercises 17–20, find the area of the triangle.
17. A = 47°,
B
c = 31
c = 28
27. a = 33.4,
b = 28.5,
c = 22.3
28. a = 18.2,
b = 17.1,
c = 12.3
29. Find the radian measure of the largest angle in the triangle with
sides of 4, 5, and 6.
30. A parallelogram has sides of 18 and 26 ft, and an angle of 39°.
Find the shorter diagonal.
37. Designing a Softball Field In softball, adjacent bases
are 60 ft apart. The distance from the center of the front edge
of the pitcher’s rubber to the far corner of home plate is 40 ft.
(a) Find the distance from the center of the pitcher’s rubber to
the far corner of first base.
(b) Find the distance from the center of the pitcher’s rubber to
the far corner of second base.
(c) Find ∠ B in ¢ABC.
6965_CH05_pp403-454.qxd
1/14/10
1:52 PM
Page 449
SECTION 5.6
Second base
c
B
Third
base
40 ft
A (First base)
60 ft
45°
C
(Home plate)
38. Surveyor’s Calculations Tony must find the distance
from A to B on opposite sides of a lake. He locates a point C
that is 860 ft from A and 175 ft from B. He measures the angle
at C to be 78°. Find distance AB.
B
175 ft
449
The Law of Cosines
42. Group Activity Architectural
Design Building Inspector Julie
F
Wang checks a building in the shape
of a regular octagon, each side 20 ft
long. She checks that the contractor
G
has located the corners of the foundation correctly by measuring several of
the diagonals. Calculate what the lengths
of diagonals HB, HC, and HD should be.
E
D
20 ft
C
20 ft
B
20 ft
H 20 ft A
43. Connecting Trigonometry
and Geometry ∠ CAB is
1 ft
inscribed in a rectangular box
A
whose sides are 1, 2, and 3 ft
long as shown. Find the measure 2 ft
of ∠ CAB.
C
3 ft
44. Group Activity Connecting
Trigonometry and Geometry
2 ft
A cube has edges of length 2 ft.
Point A is the midpoint of an edge.
Find the measure of ∠ ABC.
B
A
C
78°
860 ft
A
C
Standardized Test Questions
39. Construction Engineering A manufacturer is
designing the roof truss that is modeled in the figure shown.
(a) Find the measure of ∠CAE.
(c) Find the length EF.
C
You may use a graphing calculator when answering these questions.
47. Multiple Choice What is the area of a regular dodecagon
(12-sided figure) inscribed in a circle of radius 12?
6 ft
A
9 ft
D
E
36 ft
(A) 427
40. Navigation Two airplanes flying together in formation
take off in different directions. One flies due east at 350 mph,
and the other flies east-northeast at 380 mph. How far apart are
the two airplanes 2 hr after they separate, assuming that they
fly at the same altitude?
41. Football Kick The player waiting to receive a kickoff
stands at the 5 yard line (point A) as the ball is being kicked
65 yd up the field from the opponent’s 30 yard line. The kicked
ball travels 73 yd at an angle of 8° to the right of the receiver,
as shown in the figure (point B). Find the distance the receiver
runs to catch the ball.
Goal line
Goal line
K
45. True or False If ¢ABC is any triangle with sides and
angles labeled in the usual way, then b 2 + c2 7 2bc cos A.
Justify your answer.
46. True or False If a, b, and u are two sides and an
included angle of a parallelogram, the area of the parallelogram
is ab sin u. Justify your answer.
(b) If AF = 12 ft, find the length DF.
F
B
65 yd
0 10 20 30 40 50 40 30 20 10 0
B
48. Multiple Choice The area of a triangle with sides 7, 8,
and 9 is
(A) 6115. (B) 12 15. (C) 1613. (D) 1713. (E) 1813.
49. Multiple Choice Two boats start at the same point
and speed away along courses that form a 110° angle. If
one boat travels at 24 miles per hour and the other boat travels
at 32 miles per hour, how far apart are the boats after
30 minutes?
(A) 21 miles (B) 22 miles (C) 23 miles
(D) 24 miles (E) 25 miles
50. Multiple Choice What is the measure of the smallest
angle in a triangle with sides 12, 17, and 25?
(A) 21°
A
8°
73 yd
160
ft
(B) 432 (C) 437 (D) 442 (E) 447
(B) 22°
(C) 23°
(D) 24°
(E) 25°
Explorations
51. Find the area of a regular polygon with n sides inscribed inside
a circle of radius r. (Express your answer in terms of n and r.)
6965_CH05_pp403-454.qxd
450
1/14/10
1:52 PM
Page 450
CHAPTER 5 Analytic Trigonometry
52. (a) Prove the identity:
(a) How fast is each ship traveling? (Express your answer in
knots, which are nautical miles per hour.)
b 2 + c2 - a 2
cos A
=
.
a
2abc
(b) What is the angle of intersection of the courses of the two
ships?
(b) Prove the (tougher) identity:
2
2
2
cos A
cos B
cos C
a + b + c
+
+
=
.
a
c
b
2abc
(c) How far apart are the ships at 12:00 noon if they maintain
the same courses and speeds?
[Hint: Use the identity in part (a), along with its other
variations.]
Extending the Ideas
53. Navigation Two ships leave a
common port at 8:00 A.M. and travel
at a constant rate of speed. Each ship
keeps a log showing its distance from
port and its distance from the other
ship. Portions of the logs from later
that morning for both ships are shown
in the following tables.
Naut mi
from
Naut mi
from
Time
port
ship B
9:00
10:00
15.1
30.2
8.7
17.3
54. Prove that the area of a triangle can be found with the formula
¢ Area =
Naut mi
from
Naut mi
from
Time
port
ship A
9:00
11:00
12.4
37.2
8.7
26.0
a 2 sin B sin C
.
2 sin A
55. A segment of a circle is the region
enclosed between a chord of a circle
and the arc intercepted by the chord.
Find the area of a segment intercepted
by a 7-inch chord in a circle of radius
5 inches.
5
7
5
CHAPTER 5 Key Ideas
Properties, Theorems, and Formulas
Reciprocal Identities 404
Quotient Identities 404
Pythagorean Identities 405
Cofunction Identities 406
Odd-Even Identities 407
Sum/Difference Identities 422–424
Double-Angle Identities 428
Power-Reducing Identities 428
Half-Angle Identities 430
Law of Sines 434
Law of Cosines 442
Triangle Area 444
Heron’s Formula 445
Procedures
Strategies for Proving an Identity
413–415
CHAPTER 5 Review Exercises
Exercise numbers with a gray background indicate problems that
the authors have designed to be solved without a calculator.
The collection of exercises marked in red could be used as a chapter
test.
In Exercises 1 and 2, write the expression as the sine, cosine, or tangent of an angle.
1. 2 sin 100° cos 100°
2 tan 40°
2.
1 - tan2 40°
In Exercises 3 and 4, simplify the expression to a single term. Support
your answer graphically.
3. 11 - 2 sin2 u22 + 4 sin2 u cos2 u
4. 1 - 4 sin2 x cos2 x
In Exercises 5–22, prove the identity.
5. cos 3x = 4 cos3 x - 3 cos x
6. cos2 2x - cos2 x = sin2 x - sin2 2x
7. tan2 x - sin2 x = sin2 x tan2 x