Question

In the adjoining figure, the angle of elevation of the top $P$ of a vertical tower from a point $X$ is ${60}^0$; at a point $Y$, $40m$ vertically above $X$, the angle of elevation is ${45}^0$. Find
(i) the height of the tower $PQ$
(ii)the distance $XQ$
(Give your answer to the nearest metre)

Answer 1

Let

Height of the tower= $h$

$XQ=YR=y$

$XY=40m$

$PR=h-40$

In right $△PXQ$

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

Substituting the values,

$\Rightarrow \tan {60}^o=\frac{h}{y}$

$\Rightarrow \sqrt{3}=\frac{h}{y}$

$\Rightarrow y=\frac{h}{\sqrt{3}}...\left(1\right)$

In right triangle $PYR$

$\tan \theta =\frac{PR}{YR}$

Substituting the values,

$\Rightarrow \tan {45}^o=\frac{(h-40)}{y}$

$\Rightarrow 1=\frac{(h-40)}{y}$

$\Rightarrow y=h-40..\left(2\right)$

Using both the equations

$h-40=\frac{h}{\sqrt{3}}$

$\Rightarrow \sqrt{3}h-40\sqrt{3}=h$

$\Rightarrow \sqrt{3}h-h=40\sqrt{3}$

$\Rightarrow (1.732-1)h=40\left(1.732\right)$

$\Rightarrow 732h=69.280$

$\Rightarrow h=\frac{69.280}{0.732}=\frac{69280}{732}=94.64$

Hence, Height of the tower $=94.64m\approx 95m$

Distance $XQ=h-y=95-40=55m$.