A space shuttle pilot flying toward the Suez Canal finds that the angle of depression on one end of the canal is 38.25deg and the angle of depression to the other end is 52.75deg. If the canal is 100.6 mi long, find the altitude of the space shuttle.

Answer 1

Tan52.75 = h/d. d = hor. dist. from
shuttle to canal.
h = d*Tan52.75 = Altitude of shuttle.
Tan38.25 = h/(d+100.6)
h = (d + 100.6)* Tan38.25
Substitute d * Tan52.75 for h:
d*Tan52.75 = (d + 100.6)*Tan38.25
Solve for d:
1.3151d = 0.7883d + 79.31
1.3151d - 0.7883d = 79.31
0.5268d = 79.31
d = 79.31/0.5268 = 150.6mi = Hor.
dist. from shuttle to canal.
h = 150.6*Tan52.75 = 198mi = Alti-
tude of shuttle.