Chapter 7: Similarity

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Right
triangles
Trigonometry
7.5-7.6
DAY 1
A trigonometric ratio is a ratio of the lengths
of two sides of a right triangle.
We will use three different ratios, sine, cosine
and tangent.
Remember Chief Sohcahtoa
B
c
a
C
A
b
Trigonometric Ratio
Abbreviation
Definition
Sine of A
Sin A
opp. A
=a
Hypotenuse c
Cosine of A
Cos A
adj. to A
Hypotenuse
Tangent of A
Tan A
=b
c
opp. A = a
adj. to A b
Examples Using Sohcahtoa:
1.
A
13
5
12
sin A =
13
5
cos A =
13
tan A = 12
5
B
C
12
5
sin B =
13
12
cos B =
13
tan B = 5
12
Examples:
4. Use a calculator. Find the following,
rounding to 4 decimal places.
A) sin 27 =
.4540
B)
tan 32 = .6249
C) cos 72 = .3090
D)
sin 48 = .7431
Examples:
5.
Find the measure of the acute angles given the same trigonometric ratio.
A) sin B =
B) cos E =
15
17
4
5
C)
tan I =
12
5
I
D
A
G
62°
C
B
E
F
36.89 ~ 37°
How can I find angles with this????
2nd
sin
H
15/17
X = 67.38 ~ 67°
Tan 50 = x / 4.8
angles of
elevation and
Depression
DAY 2
7-7 Angles of Elevation and Depression
• By definition, the picture at the left shows each of the
angle of depression and the angle of elevation.
However, since the angles are congruent because
the lines are ll, then Alternate Interior angles are congruent
___________________________________,
we can just
use the triangle at the right
Class Exercises:
1.
A surveyor is 130 ft. from a tower. The tower is 86 ft. high. The
surveyor’s instrument is 4.75 feet above the ground. Find the angle
of elevation.
Tan x = (86-4.75)/130
Tan x = 81.25/130
Angle of elevation = ??
130 ft
Tan x = .625
X = 32°
86 ft
4.75 ft
2. A plane P is 3 miles above ground. The pilot sights the airport A at an
angle of depression of 15o. He sights his house H at an angle of
depression of 32o. What is the ground distance d between the pilot’s
house and the airport.
Airport distance
Home distance
Tan 15 = 3/a
Tan 32 = 3/h
a = 3/tan 15
h = 3/tan 32
a = 11.2 miles
h = 4.8 miles
Airport – home = 11.2 - 4.8 =6.4 miles
32
15
15
32
3 miles
Class Exercises:
3. State an equation that would enable you
to solve each problem. Then solve.
Round answers to the nearest tenth.
a. Given mP = 15 and PQ = 37, find QR.
Sin 15 = x/37
37 * sin 15 = x
9.6 = x
b. Given PR = 2.3 and PQ = 5.5, find mP.
Cos P = 2.3/5.5
Cos P = .41818
P = 65.3
Q
R
P
Class Exercises:
4.
A truck is driven onto a ramp that is 80 ft long. How high is the
end of the ramp when the angle of elevation of the ramp is 30 ?
45  ?
Sin 30 = x/80
80 * sin 30 = x
40 Ft = x
Sin 45 = x/80
80 ft
X ft
30
80 * sin 45 = x
56.6 Ft = x
Class Exercises:
5. A 6 ft tall person is
enjoying a Saturday
afternoon by flying a kite.
The angle of elevation from
him to the kite is 35 . He
brought 75 ft of string and
has used all of it. How high
is the kite?
75 ft
x ft
35
Sin 35 = x/75
75 * sin 35 = x
43 ft = x
The kite is 43 + 6 = 49 ft above the ground.
6 ft
System of equations with trig
• You are in a building on the 3rd floor and look across the street at a
crane. You notice that the angle when you look up at the top of the
crane is 55° and the angle when you look down to the bottom is 61°.
If the crane is 60 ft tall, how far away is the building you are in ?
(60-y)
60ft
x
55°
61°
y
Step 1: Write the 2 equations for the
system.
Step 2: Solve each equation for y=.
Step 3: Set the equations equal to each
other and solve for x.
Step 4: If you need the answer for y
plug in the value for x and solve for y.
x tan 61   x tan 55  60
y
tan 61 
x
y  x tan 61
(60  y )
tan 55 
x
x tan 61  x tan 55  60
x(tan 61  tan 55)  60
x tan 55  60  y
60
x
(tan 61  tan 55)
x tan 55  60   y
x  18.563
ft away from the crane
y   x tan 55  60

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