Question

The angle of elevation $\theta$ of a vertical tower from a point on the ground is such that its tangent is $\left(\frac{5}{12}\right)$. On walking 192 metres towards the tower in the same straight line, the tangent of the angle of elevation $\Phi$ is found to be $\left(\frac{3}{4}\right)$. The height of the tower is :

• A. 200 m
• B. 160 m
• C. 220 m
• D. 180 m

Answer 1

The correct option is D 180 m

Let PQ be the tower and let A and B be the given points of observation on the ground. Then,
$\angle PAQ=\theta ,\angle PBQ=\Phi ,\angle APQ={90}^{\circ },AB=192m,\tan \theta =\frac{5}{12}$ and $\tan \Phi =\frac{3}{4}$
Let PQ = h metres and BP = x metres.

From right $\Delta APQ$, we have

$\tan \theta =\frac{PQ}{AP}$

$\Rightarrow \frac{h}{192+x}=\frac{5}{12}$

$\Rightarrow 5x=(12h-960)$

$\Rightarrow x=\frac{12h-960}{5}\dots \left(1\right)$

From right $\Delta BPQ$, we have

$\tan \Phi =\frac{PQ}{BP}$

$\Rightarrow \frac{h}{x}=\frac{3}{4}$

$\Rightarrow 3x=4h$

$\Rightarrow x=\frac{4h}{3}\dots \left(2\right)$

From (1) and (2), we get

$\frac{12h-960}{5}=\frac{4h}{3}$

$\Rightarrow 36h-2880=20h$

$\Rightarrow 16h=2880$

$\Rightarrow h=180m$

Hence, the height of the tower $=180m$