Question

From a window ($h$ metres high above the ground) of a house in a street, the angle if elevation and depression of the top and the foot of another house on the opposite side of the street are $\theta$ and $\varphi$ respectively.
Find the height of opposite house if $\theta ={60}^o,\varphi ={45}^o,h=120$metres.

• A. $317.84m$
• B. $366.97m$
• C. $301.44m$
• D. None of these

Answer 1

The correct option is D None of these

Let $L$ be the window of a house $LM$. Let $PQ$ be the height of the opposite house such that, $PQ=$ height of the opposite house $=H$ metres. Then $LM=h$metres. Through $L$ draw $LN\perp PQ$, meeting $PQ$ at $N$. As per condition of the problem, $\angle PLN=\theta ,\angle NLQ=\varphi$, Let $PQ=H$ metres then $PN=(H-h)$metres. Let $LN=MQ=x$metres
($\because NQ=LM=h$metres)
From the right $△LNP$,
$\frac{PN}{LN}=\tan \theta$
$\frac{H-h}{x}=\tan \theta$
$x=\frac{H-h}{\tan \theta }.......\left(1\right)$
From the right $△LNQ$
$\frac{NQ}{LN}=\tan \varphi$
$\frac{h}{x}=\tan \varphi$
$\frac{h}{\tan \varphi }=x.........\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get:
$\frac{H-h}{\tan \theta }=\frac{h}{\tan \varphi }$
$\Rightarrow$ $H\tan \varphi -h\tan \varphi =h\tan \theta$
$\Rightarrow$ $H\tan \varphi =h(\tan \varphi +\tan \theta )$
Required height of opposite house.
$=H=h(1+\tan \theta \cot \varphi )$
For $\theta ={60}^o,\varphi ={45}^o,h=120m$,
The height of the opposite house
$H=120(1+\tan {60}^o\cot {45}^o=120(1+\sqrt{3}\times 1)$
$H=120(1+1.732)=120\times 2.732=327.84m$