Transcript 4.8 Applications and Models

4.8
Applications and
Models
Solve the right triangle.
B
B = ? = 90 – 34.2 = 55.8
c
C
a
tan 34.2 
a = 13.18
19.4

a
34.2
b = 19.4
19.4
A cos 34.2 
c
 23.46
h 40’
If the ladder is 40’ long
and the angle of elevation
is 42 , how high up is the
fireman?
42
h
sin 42 
40
 26.77 ft.
At a point 200 feet from the base of a building, the angle
of elevation to the bottom of a smokestack is 35, while
the angle to the top is 53. Find the height, s, of the
smokestack alone.
s
x
tan 35 
200
xs
tan 53 
200
53
x
35
200 ft.
x  140.04
x  s  265.41
265.41
-140.04
125.37 ft.
Trigonometry and Bearings
N
N
80
W
E
S
35
S 35 E
W
E
S
N 80 W
A ship leaves port at noon and heads due west at 20
knots (nautical miles per hour). At 2 P.M. , to avoid
a storm, the ship changes course to N 54 W. Find the
ship’s bearing and distance from the port of departure
at 3 P.M..
Now find angle A.
11.76
tan A 
16.18  40
D
c
b =11.76
20 nm
54
36
d = 16.18
B
 11.82
N 78.18W
 11.82
2(20) = 40 nm
A
Find c.
c = 57.4 nm