Question 1093161
Let x = horizontal distance of the plane from the telescope when the angle of elevation is θ. Then:
x/5 = cotθ
(1/5) dx/dt = - cosec^2 θ * dθ/dt
So, the horizontal speed at which the plane is flying is:
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