Transcript ITrig 6.7 - Souderton Math

Applications and Models
Objective: To solve a variety of
problems using a right triangle.
Example 1
• Solve the right triangle for
all missing sides and angles.
Example 1
• Solve the right triangle for
all missing sides and angles.
• The angles of the triangle
need to add to 1800.
1800 – 900 – 34.20 = 55.80
Example 1
• Solve the right triangle for
all missing sides and angles.
a
tan 34.2 
19.4
0
19.4 tan 34.2 0  a
13.18  a
55.80
Example 1
• Solve the right triangle for
all missing sides and angles.
a
tan 34.2 
19.4
0
19.4
cos 34.2 
c
0
19.4 tan 34.2 0  a
13.18  a
19.4
c
cos 34.2 0
c  23.46
55.80
You Try
• Solve the right triangle for
all missing sides and angles.
You Try
• Solve the right triangle for
all missing sides and angles.
• The missing angle is
1800  900  280  620
You Try
• Solve the right triangle for
all missing sides and angles.
40
tan 28 
b
0
40
b
tan 280
b  75.23
62 0
You Try
• Solve the right triangle for
all missing sides and angles.
40
tan 28 
b
0
40
b
tan 280
b  75.23
40
sin 28 
c
0
40
c
sin 280
c  85.20
62 0
Example 2
• A safety regulation states that the maximum angle of
elevation for a rescue ladder is 720. A fire
department’s longest ladder is 110 feet. What is the
maximum safe rescue height?
Example 2
• A safety regulation states that the maximum angle of
elevation for a rescue ladder is 720. A fire
department’s longest ladder is 110 feet. What is the
maximum safe rescue height?
• We need to solve for a.
a
0
sin 72 
110
110 sin 720  a
104.6  a
Example 3
• At a point 200 feet from the base of a building, the
angle of elevation to the bottom of a smokestack is
350, whereas the angle of elevation to the top is 530.
Find the height s of the smokestack alone.
Example 3
• At a point 200 feet from the base of a building, the
angle of elevation to the bottom of a smokestack is
350, whereas the angle of elevation to the top is 530.
Find the height s of the smokestack alone.
• First, we find the height to
the top of the smokestack.
as
tan 53 
200
0
200 tan 530  a  s
265.41  a  s
Example 3
• At a point 200 feet from the base of a building, the
angle of elevation to the bottom of a smokestack is
350, whereas the angle of elevation to the top is 530.
Find the height s of the smokestack alone.
• Next, we find the height
of the building.
a
tan 35 
200
0
200 tan 350  a
140.04  a
Example 3
• At a point 200 feet from the base of a building, the
angle of elevation to the bottom of a smokestack is
350, whereas the angle of elevation to the top is 530.
Find the height s of the smokestack alone.
• We subtract the two
values to find the height
of the smokestack.
265.41 140.04  125.37
Example 4
• A swimming pool is 20 meters long and 12 meters
wide. The bottom of the pool is slanted so that the
water depth is 1.3 meters at the shallow end and 4
meters at the deep end, as shown. Find the angle of
depression of the bottom of the pool.
Example 4
• A swimming pool is 20 meters long and 12 meters
wide. The bottom of the pool is slanted so that the
water depth is 1.3 meters at the shallow end and 4
meters at the deep end, as shown. Find the angle of
depression of the bottom of the pool.
2.7
tan  
20
 2.7 
tan 
 
 20 
1
7.69  
Trig and Bearings
• In surveying and navigation, directions are generally
given in terms of bearings. A bearing measures the
acute angle that a path or line of sight makes with a
fixed north-south line. Look at the following
examples.
Trig and Bearings
• In surveying and navigation, directions are generally
given in terms of bearings. A bearing measures the
acute angle that a path or line of sight makes with a
fixed north-south line. Look at the following
examples.
Trig and Bearings
• You try. Draw a bearing of:
W200N
E300S
Trig and Bearings
• You try. Draw a bearing of:
W200N
E300S
20 0
300
Class work
• Pages 521-522
• 2-8 even
Homework
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Pages 521-522
1-7 odd
15-21 odd
27, 29, 31