Geometry Chapter 7 Notes

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Chapter 7
Right trianlges and
Trigonometry
SECTION 7.1:
Geometric Mean
A Geometric Mean is the multiplicative
average between two numbers.
The Geometric Mean, x, is
proportionally between the numbers a
and b.
 
=
 
Find the Geometric Mean between each pair of
numbers.
1. 4 and 9
2. 6 and 15
Find the Geometric Mean between each pair of
numbers.
1. 4 and 9
2. 6 and 15
4

1.
=

9
4(9) = x(x)
36 = x2
36 =  2
6=x
6

2. =

15
6(15) = x(x)
90 = x2
90 =  2
9.49 = x
When the Altitude of a Right Triangle is drawn from
the Right Angle to the Hypotenuse, it divides the
Triangle into 3 Similar Triangles…
∆ACD ~ ∆CBD ~ ∆ABC
Baby ~ Momma ~ Poppa
B
B
C
A
D
C
D
A
C
Therefore there are 3 different
Geometric Means formed
• #1 The altitude splits the hypotenuse, h is the GM between x and y


=


• #2 & #3 The legs each connect twice to the hypotenuse,
• b is the GM between x and c
• a is the GM between y and c


=




=


In ∆PQR, RS = 3 m and QS = 14 m. Find PS.
In ∆PQR, RS = 3 m and QS = 14 m. Find PS.
PS is the GM between RS and QS.
Let PS = x.


=


42 = x2
 = 
x ≈ 6.48, PS ≈ 6.48 m
Find x and y in ∆PQR.
Find x and y in ∆PQR.
SECTION 7.2:
The Pythagorean
Theorem and its
Converse
Solve for x.
Solve for x.
inches
Groups of INTEGERS that form Right Triangles
are called PYTHAGOREAN TRIPLES
• 3, 4, 5
• 5, 12, 13
• 7, 24, 25
• 8, 15, 17
• 9, 40, 41
• 11, 60, 61
You can use them with
multipliers too….
3, 4, 5 doubled is 6, 8, 10
• They’re worth remembering…
Determine if ∆PQR is a right Triangle.
Determine if ∆PQR is a right Triangle.
Since the sum of the
squares of the shorter
two sides is equal to
the square of the
longest side, ∆PQR is
a right triangle.
SECTION 7.3:
Special Right
Triangles
In your groups, solve each triangle for “x”.
a.
b.
x
1
1
c.
x
2
2
x
3
3
Answer the following questions about the ∆s:
• What type of triangle are
these? Be as specific as
possible.
• What are the three angles
measures of the triangles?
• What pattern do you see?
• Are there any conclusions
your group can draw about
this type of triangle?
A
∆ABC is an equilateral
triangle with side
length of 2x units.
1
 is an Altitude and Median.
Find:
• The value of a
• The value of b
• The measure of ∠1
• The measure of ∠2
• The measure of ∠3
b
C
2
3
D
a
B
A
 is an Altitude and Median.
Find:
• The value of a = x
•
•
•
•
1
The value of b = x 3
The measure of ∠1 =30°
The measure of ∠2 =60°
The measure of ∠3 =90°
b
C
2
3
D
a
B
45° - 45°- 90° Triangles
Angle
Measure
45° 45° 90°
Side
Across
n
n
 2
Find x.
Find x.
30° - 60°- 90° Triangles
Angle
Measure
Side
Across
30° 60° 90°
n n 3 2n
Find the length of segment AC.
Find the length of segment AC.
Angle
Measure
30°
60°
90°
Side
Across
n
n 3
2n
AB
AC
BC
Find the length of segment AC.
Angle
Measure
30°
Side
Across
n
60°
n 3
90°
2n
in
SECTION 7.4:
Trigonometric
Ratios
Trigonometric Ratios define relationships between
1 Angle and 2 Sides of a Right Triangle:
• The Sine of an angle is the ratio of Opposite Side to
Hypotenuse
• The Cosine of an angle is the ratio of Adjacent Side to
Hypotenuse
• The Tangent of an angle is the ratio of Opposite Side
to Adjacent
Signs Of Happiness - Come After Having Taken On Algebra
•Soh
•Cah
•Toa

•sin Ꝋ =
ℎ

•cos Ꝋ =
ℎ

•tan Ꝋ =

The “sides” depend on the angle:
• Angle A
• Angle B
B
B
5
5
A
A
C
C
4
4
• sin A =
3
3
3
5
• tan A =
cos A =
3


5
• sin B =
4
5
• tan A =
cos B =
4


5
Steps for Solving Soh Cah Toa Problems:
• 1. Identify the Angle
• 2. Label/Identify the Sides (hypotenuse,
opposite, adjacent)
• 3. Match with Soh, Cah, Toa
• 4. Plug in values
• 5. Solve (3 methods – multiply, switch or inverse)
Method 1: “x” on top… MULTIPLY
B
10
x
A
41°
C
Method 1: “x” on top… MULTIPLY
B
10
x
A
41°
C
• Angle is A, 41°
• Hypotenuse = 10, Opposite = x
• Soh, Cah, Toa…. H and O = SINE

sin 41° =


10(sin 41° ) = ( )10

10(sin 41° ) = x ≈ 6.56 u
Method 2: “x” on bottom… SWITCH
B
x
62°
12
A
C
Method 2: “x” on bottom… SWITCH
B
x
62°
12
• Angle is B, 62°
• Hypotenuse = x, Adjacent = 12
• Soh, Cah, Toa…. H and A = COS
cos 62° =


SWITCH!
A
C
x=

cos 62°
x ≈ 25.56 u
Method 3: “x” is the angle… INVERSE
B
11
A
x°
C
18
Method 3: “x” is the angle… INVERSE
B
11
• Angle is A, x°
• Opposite = 11, Adjacent = 18
• Soh, Cah, Toa…. O and A = TAN
tan x° =
A
x°
C
18


INVERSE! 2nd Button

−
(tan x°) =
− 
 ( )

x ≈ 31.43°
SECTION 7.5:
Angles of
Elevation and
Depression
Angles of Elevation and Depression
•Angles formed by “line of sight” and a
HORIZONTAL line
•Elevation is “looking up”
•Depression is “looking down”
•∠Elevation ≅ ∠Depression
by “Z-Rule”
•Just SOH CAH TOA Word Problems
Mrs. Evans is standing on cliff, 40 meters above the
ground. She sees a group of students in need of tutoring,
at an angle of depression of 34°. How far from the base of
the cliff are they standing?
Since the angle of depression is equal to the angle of elevation, we can use either. The
angle of elevation is easier to use, since the “triangle” is clear. Once you’ve drawn the
diagram and labelled the figure, then just use our steps for Soh Cah Toa problems.
Angle is 34°
Opposite side = 40, Adjacent Side = x
Soh, Cah, Toa…. O and A = TANGENT
tan 34° =



x=
tan 34°
x ≈ 59.30 m
SECTION 7.6:
Law of Sines
Find b. Round to the nearest hundredth.
Find b. Round to the nearest hundredth.
b ≈ 4.62 u
• Draw a picture to help:
• Draw a picture to help:
We need to find C, a, and c to
Solve the Triangle. C is easy, using
the 180° Rule.
33 + 47 = 80. 180 – 80 = 100.
Therefore, ∠C = 100°
Finding sides a and c, use the Law
of Sines.
SECTION 7.7:
Law of Cosines
Solve for a side if you know 2
sides and the angle between
them:
 =  +  −   
Or solve for an angle if you
know all three sides:
 +  − 
  =

a ≈ 9.17 u
 +  − 
  =

 +  − 
  =
  ()

  =

 = 
−

(
)

R ≈ 29.09°
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