Question

A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is a/4, this angle is decreasing at a rate of 1/3 rad/min. How fast is the

plane traveling at that time?

Answer 1

-csc^2 θ = dθ/dt = 1/5 (dx/dt)

or

3.49 km/min

Explanation:

dx/dt

- 5cosec^2 θ * (rate of change of angle of elevation)

- 5 cosec^2 (π/3) * (-π/6)

(5π/6) * (4/3)

= 10π/9 km/min. = 3.49 km/min