Question

A house of height 100 m subtends a right angle at the window of an opposite house. If the height of the window is 64 m, then the distance between the two houses is

• A. 48 m
• B. 36 m
• C. 54 m
• D. 72 m

Answer 1

Answer: (A)

48 m

Let $h_1$ and $h_2$ be the two houses and $d$ be the distance between them.

Given that, height of the $h_1$ is $100m$ and the height of the window of $h_2$ is $64m$.

$h_1$ subtends a right angle $\left({90}^o\right)$ at the window of $h_2$.

Based on the given information, consider the above figure.

In $\Delta BCD$,

$\tan \theta =\frac{64}{d}....\left(i\right)$ $[\because \tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}]$

Also, in $\Delta AED$,
$\tan \theta =\frac{64}{100-64}$

$\Rightarrow (100-64)\tan \theta =d$
Substituting value of $\tan \theta$ from $\left(i\right)$, we get:

$\Rightarrow 36\left(\frac{64}{d}\right)=d$

$\Rightarrow d^2=36\times 64$

$\Rightarrow d=6\times 8=48m$

Hence, the distance between the two houses is $48m$.