Question

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $\frac{5}{12}$. On walking 192 metres towards the tower; the tangent of the angle is found to be $\frac{3}{4}$. Find the height of the tower.

Answer 1

First case:-
$tan\alpha =\frac{5}{12}$=\frac{ height of tower}{distance} (d) ---------------(1)

second case:-
$\frac{3}{4}=\frac{heightofthetower}{(d-192)}$-------(2)

from equation (1) and (2)

$3\frac{(d-192)}{4}=\frac{5d}{12}$
$9(d-192)=5d$
9d - 1728=5d
4d=1728
d=432 m

Now
$heightoftower=5\times \frac{432}{12}=5\times x\times 36=180m$