Download 5.6 Right Triangle Trigonometry

5.6-Right Triangle Trigonometry
Example: From the top of a lighthouse 85 feet high, the angle of depression to a small boat is 51.6
degrees. How far from the foot of the lighthouse is the boat?
Solution: The angle of depression is 51.6 degrees. Subtract this from 90 to obtain 38.4 degrees.
x
85
x = 85 tan (38.4)
x = 85(0.7925)
tan 38.4 =
x = 67.370
x ≈ 67.4
The boat is approximately 67 feet from the lighthouse.
Example: An airplane is on short final at 200 feet above the ground (AGL) and the angle of depression
to the landing point is 30 degrees. How far is the landing point from the airplane?
Solution: The angle of depression is 30 degrees. Subtract this from 90 to obtain 60 degrees.
200
x
200
x=
cos 60
200
x=
0.5
x = 400
cos 60 =
The aircraft is 400 feet from the landing point.
Example: A certain rock rises almost straight upward from the valley floor. From one point, the angle
of elevation of the top of the rock is 17.8 degrees. From a point 134 meters closer to the rock, the angle
of elevation of the top of the rock is 32.1 degrees. How high is the rock?
Solution: Draw a picture representing the situation.
Create trigonometric relationships between the given values.
tan 17.8 =
H
x + 134
and
tan 32.1 =
H
x
We now have a system of two equations in two variables. The easiest way to solve this system will be
by substitution. Begin by solving both equations for H.
H = ( x + 134) tan 17.8
and
H = x tan 32.1
Set the equations equal to each other.
x tan 32.1 = ( x + 134) tan 17.8
x tan 32.1 = x tan 17.8 + 134 tan 17.8
x tan 32.1 − x tan 17.8 = 134 tan 17.8
x(tan 32.1 − tan 17.8) = 134 tan 17.8
134 tan 17.8
x=
tan 32.1 − tan 17.8
43.022
x=
0.627 − 0.321
x = 140.529
Therefore,
H = x tan 32.1
H = 140.529 tan 32.1
H = 88.154
H ≈ 88.2
The rock is approximately 88.2 meters high.
Example: A hot air balloon is rising upward from the earth at a constant rate. An observer 248 meters
away spots the balloon at an angle of elevation of 27 degrees Three minutes later the angle of elevation
of the balloon is 62 degrees. At what rate is the balloon ascending in meters per second?
Solution: Draw a picture representing the situation.
Find H1:
H1
248
H 1 = 248 tan 27
tan 27 =
H 1 = 248(0.5095)
H 1 = 126.362
Find H2:
H2
248
H 2 = 248 tan 62
tan 62 =
H 2 = 248(1.880)
H 2 = 466.42
The difference between H1 and H2 is the distance the balloon rose in three minutes.
H 2 − H1 = 466.420 − 126.362 = 340.06
The balloon rose 340.06 meters in three minutes or a rate of 113.3 meters per minute. To convert to
meters/sec., divide this by 60.
340.06meters 1 min .
∗
= 1.889
3 min
60 sec .
The balloon is rising at a rate of 1.9 meters/sec.
Example: An airplane takes off from the Deer Valley airport and climbs out at an angle of 50 degrees
from the horizontal. What is the aircrafts altitude to the nearest foot after it has travelled a distance of
1,500 feet? What is the airplanes climb rate if it took 20 seconds to climb to this altitude?
Solution: Draw a picture representing the situation.
Let the altitude be equal to x.
x
1500
x = 1500 sin 50
sin 50 =
x = 1500(0.766)
x = 1149.06
The airplane is at an altitude of approximately 1,149 feet after it has travelled 1500 feet. Because it took
20 seconds to gain 1,149 feet of altitude its climb rate is:
1149 60 sec .
∗
= 3447
20 sec . 1 min
The airplane is climbing at a rate of 3,447 feet/ min.
Example: An airplane takes off from the Deer Valley airport and climbs out at an angle of 55 degrees
from the horizontal. What is the aircrafts altitude after it has travelled a distance of 1,000 feet? What is
the airplanes climb rate if it took 15 seconds to climb to this altitude?
Solution: Draw a picture representing the situation.
Let the altitude be equal to x.
x
sin 55 =
1000
x = 1000 sin 55
x = 1500(0.8191)
x = 819.15
The airplane is at an altitude of approximately 819 feet after it has travelled 1000 feet.
Because it took 15 seconds to gain 819 feet of altitude its climb rate is:
819
60 sec .
∗
= 3276.6
15 sec . 1 min
The airplane is climbing at a rate of 3,277 feet/ min.