Transcript bYTEBoss Lesson 12.2 Applications and Angles of Elevation and

Chapter 12 - Lesson 12.2
Problem Solving with Right Triangles
HW: 12.2/1-20
Finding the Sides of a Triangle
Remember:
SOHCAHTOA
O A O
S C T
H H A
Review: Trig Ratios
First we will find the Sine, Cosine and Tangent
ratios for Angle P.
P
Next we will find the Sine,
Cosine, and Tangent ratios for
Angle Q
16
20
Sin P

Cos P
12

20
Tan P
16

12
12
Adjacent
20
16
Opposite
Sin Q

12
20
Cos Q

16
20
Tan Q

12
16
Remember SohCahToa
Q
Solving Right Triangles
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.
Missing sides r and s,
and angle S.
 S  90  20
 S  70
r
sin 20 
15
s
cos 20 
15
15sin 20  r 15cos 20  s
r  5.1303
r  5.1
s  14.0954
s  14.1
But what if you don’t know either of the acute angles?
To solve those triangle we must use
Inverse Trig Functions
10
tan  A 
 1.25
8
 A  tan 1.25
1
 A  51.3
 B  90  51.3  38.7
b  8  10
2
2
2
b  64  100
b  164  12.8
Missing
side….
10 5
cos B 

12 6
cos B  0.8
B  cos 0.8
B  36.869  36.9
1
12  10  b
2
2
2
b  144  100  44
b  6.6332  6.6
m  A  90  36.9
m  A  53.1
What are the angles of elevation and
depression and what is their
relationship to right triangles?
ANGLE OF DEPRESSION
Looking down from the horizontal
observer
Eye level
Angle of
depression
cliffs
object
Sea level
ANGLE OF ELEVATION
Looking up from the horizontal
observer
Eye level
cliffs
Angle of
elevation
object
Sea level
If an observer sights an object above, the
angle between a horizontal line and his or
her line of sight is called an angle of
elevation. If the line of sight is below the
horizontal it is called the angle of
depression.
Angle of
Depression
Angle of Elevation
eye - level
eye - level
eye - level
eye - level
The angles are equal –
they are alternate interior angles
eye - level
eye - level
Angles of Elevation and Depression
Top Horizontal
Angle of Depression
Angle of Elevation
Bottom Horizontal
Since the two horizontal lines are parallel, by Alternate Interior Angles
the angle of depression must be equal to the angle of elevation.
Angles of Elevation and Depression
Example 1
The angle of elevation of building A to building B is 250. The
distance between the buildings is 21 meters. Calculate how much
taller Building B is than building A.
Step 1: Draw a right
angled triangle with
the given information.
Step 2: Take care with
placement of the angle
of elevation
Step 3: Set up the
trig equation.
Step 4: Solve the trig
equation.
Angle of
elevation
B
h
A
250
21 m
tan 25 
h
21
h  21 tan 25
h  9.8 m
Example 2 A boat is 60 meters out to sea. Madge is standing on a
cliff 80 meters high. What is the angle of depression
from the top of the cliff to the boat?
Step 1: Draw a right
angled triangle with the
given information.
Step 2: Alternate interior
angles place  inside the
triangle.
Step 3: Decide which
trig ratio to use.
Step 4: Use calculator to
find the value of the
unknown.

Angle of
depression
80 m

60 m
80
tan  
60
 80 
  tan 1  
 60 
  53.1º
Example 3
Marty is standing on level ground when he sees a plane directly
overhead. The angle of elevation of the plane after it has travelled
25 km is 200. Calculate the altitude of the plane at this time.
Step 1: Draw a right angled
triangle with the given
information.
Step 2: Alternate interior angles
places 200 inside the triangle.
Step 3: Decide which trig ratio
to use.
Step 4: Use calculator to find
the value of the unknown.
Plane
25 km
200
h
Angle of
elevation
200
h
tan 20 
25
0
(nearest km)
Examples
Example 4
Kate and Petra are on opposite sides of a tree. The angle of
elevation to the top of the tree from Kate is 45o and from Petra is
65o. If the tree is 5 m tall, who is closer to the tree, Kate or Petra?
5m
K
450
k
Kate
tan 45 
k
p
P
Petra
5
k
5
tan 45
k  5m
650
tan 65 
p
5
p
5
tan 65
p  2.3 m (2 sig . figs )
Answer
Therefore, Petra
is closer to the
tree, since the
distance is
shorter.
Example 5
Maryann is peering outside her window. From her window she
sees her car and a bird hovering above her car. The angle of
depression of Maryann’s car is 200 whilst the angle of elevation to
the bird is 400. If Maryann’s window is 2m off the ground, what is
the bird’s altitude at that moment?
Step 1: Draw a diagram
Bird
400
200
b
d
2m
Car
Therefore,
Step 2: Set up the trig equations in
two parts. Find d first, then b.
Step 3: Solve the equations and answer
the question.
2
tan 20 
d
b
tan 40 
5 .5
2
b  5.5 tan 40
d
tan 20
b  4 .6 m
d  5.5 m
The bird is 6.6 m (2 + 4.6) from the ground at that moment.
Your Turn 1:
You sight a rock climber on a cliff at a 32o angle of
elevation. The horizontal ground distance to the cliff is
1000 ft. Find the line of sight distance to the rock
climber.
1000
Cos 32 
x
1000
x
Cos 32
x
x  1179 ft
32
1000 ft
Your Turn 2:
An airplane pilots sights a life raft at a 26o angle of
depression. The airplane’s altitude is 3 km. What is
the airplane’s surface distance d from the raft?
3
Tan 26 
d
3
d
Tan 26
26
3 km
26
d  6.2 km
d
FYI: Theodolite
Theodolite is a precision instrument for measuring angles in the
horizontal and vertical planes.
When the telescope is
pointed at a target
object, the angle of each
of these axes can be
measured with great
precision
Theodolites are still
used today for ultra
high precision optical
alignment and
measurement
Theodolites are mainly
used for surveying
applications, and have
been adapted for
specialized purposes in
fields like meteorology
and rocket launch
technology.
Your Turn 3:
A surveyor stands 200 ft from a building to measure its height with
a 5-ft tall theodolite. The angle of elevation to the top of the
building is 35°. How tall is the building?
tan 35° =
x
200
x = 200 • tan 35°
So x ≈ 140
To find the height of the building, add the height of the Theodolite,
which is 5 ft tall.
The building is about 140 ft + 5 ft, or 145 ft tall.
Your Turn 4:
An airplane flying 3500 ft above ground begins a 2° descent to
land at an airport. How many miles from the airport is the airplane
when it starts its descent?
sin 2° = 3500
x ft
x ft = 3500 ft
sin 2°
x ft ≈ 100,287.9792 ft
100287.9792 ft  5280 ft / mile
x miles ≈ 18.9939 miles ≈ 19 miles
The airplane is about 19 mi from the airport when it
starts its descent.
Your Turn 5:
A 6-ft man stands 12 ft from the
base of a tree. The angle of
elevation from his eyes to the
top of the tree is 40°.
1. About how tall is the tree?
about 16 ft
2. If the man releases a pigeon that flies directly to the
top of the tree, about how far will it fly?
about 15.7 ft
3. What is the angle of depression from the treetop to
the man’s eyes?
40°
Your Turn 6:
CIRCUS ACTS. At the circus, a person in the audience
watches the high-wire routine. A 5-foot-6-inch tall acrobat is
standing on a platform that is 25 feet off the ground. How
far is the audience member from the base of the platform, if
the angle of elevation from the audience member’s line of
sight to the top of the acrobat is 27°?
Step 1: Draw a triangle to fit problem
30.5 = 25 + 5.5
opp
Step 2: Label sides from angle’s view
Step 3: Identify trig function to use
27°
x adj
Step 4: Set up equation
Step 5: Solve for variable
SO/H
CA/H
TO/A
30.5
tan 27° = ------x
x tan 27° = 30.5
x = (30.5) / (tan 27°)
x = 59.9
Your Turn 7:
DIVING At a diving competition, a 6-foot-tall diver stands
atop the 32-foot platform. The front edge of the platform
projects 5 feet beyond the ends of the pool. The pool itself is
50 feet in length. A camera is set up at the opposite end of
the pool even with the pool’s edge. If the camera is angled so
that its line of sight extends to the top of the diver’s head,
what is the camera’s angle of elevation to the nearest degree?
37
tan x 
45
 37 
x  tan  
 45 
37 = 32 + 6
1
x  39.427  39
x°
45 = 50 - 5
Your Turn 8:
From a point 80m from the base of a tower, the angle of
elevation is 28˚. How tall is the tower to the nearest meter?
x
28˚
80m
x
tan 28˚ = 80
80 (tan 28˚) = x
80 (.5317) = x
x ≈ 42.5m is the height of the tower
Your Turn 8:
A ladder that is 20 ft is leaning against the side of a building.
If the angle formed between the ladder and ground is 75˚,
how far is the bottom of the ladder from the base of the
building?
building
20
75˚
x
x
cos 75˚ = 20
20 (cos 75˚) = x
20 (.2588) = x
x ≈ 5.2ft from the base of the building
Your Turn 9:
When the sun is 62˚ above the horizon, a building casts a
shadow 18m long. How tall is the building?
x
62˚
18 shadow
x
tan 62˚ = 18
18 (tan 62˚) = x
18 (1.8807) = x
x ≈ 33.9m is the height of the building
Your Turn 10:
A kite is flying at an angle of elevation of about 55˚. Ignoring
the sag in the string, find the height of the kite if 85m of
string have been let out.
kite
85
x
55˚
x
sin 55˚ = 85
85 (sin 55˚) = x
85 (.8192) = x
x ≈ 69.6m is the height of the kite
Your Turn 11:
A 5.5 foot person standing 10 feet from a street light casts a
14 foot shadow. What is the height of the streetlight?
5.5
10
1st find the angle of elevation
tan x˚ =
5.5
14
x° ≈ 21.45°
14
shadow
x˚
2nd use the angle to find the
height of the light.
height
tan 21.45 
24
height  24tan 21.45
height = 9.43 feet
Your Turn 12:
The angle of depression from the top of a tower to a boulder
on the ground is 38º. If the tower is 25m high, how far from
the base of the tower is the boulder?
38º
angle of depression
Alternate Interior Angles are congruent
25
38º
x
25
tan 38 
x
x
25
tan 38
x  31.99  32m