A man is moving away from a tower 41.6 m high at the rate of 2 m/sec. Find the rate at which the angle of elevation of the top of tower is changing, when he is at a distance of 30m from the foot of the tower. Assume that the eye level of the man is 1.6m from the ground.

Explanation

`tantheta=(BC)/(PC)=40/x`

`x=40cottheta`

`dx/dt=-40cosec^2theta(d theta)/dt`

`dx/dt=2m|sec`

`(d theta)/(dt)=-1/(20cosec^2theta)`

When`x=30`

`coththeta=30/40=3/4`

`cosecc^2theta=1+cot^2theta`

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