Question

An airplane at a constant altitude of 2 miles flies a horizontal distance toward you at a constant velocity. At the start of your observation, the angle of elevation 40°. Fifteen seconds later, the angle of elevation is 50°. What the approximate velocity of the airplane in miles per minute?

Answer 1

distance traveled = 2(cot40°-cot50°) = 0.7053 miles
so the speed is 2.82 mi/min

Answer 2

Let x be the horizontal distance in miles that the airplane flies towards you and let v be its velocity in miles per minute. Then we have:

tan(40) = 2/x (at the start of observation)
tan(50) = 2/(x+15v) (15 seconds later)

Multiplying the first equation by (x+15v) and simplifying, we get:

2tan(40)x + 2tan(40)*15v = 2(x+15v)tan(50)

Substituting the values of tan(40) and tan(50), we get:

2x/0.839 + 15v/0.839 = 2(x+15v)/1.1918

Simplifying, we get:

2.383x + 12.259v = 2.114x + 31.718v

Solving for v, we get:

v ≈ 2.75 miles per minute

Therefore, the approximate velocity of the airplane is 2.75 miles per minute.