Question

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

• A. $\mathbf{48}\mathbf{}\mathit{m}$
• B. $\mathbf{36}\mathbf{}\mathit{m}$
• C. $\mathbf{54}\mathbf{}\mathit{m}$
• D. $\mathbf{72}\mathbf{}\mathit{m}$

Answer 1

The correct option is A

$\mathbf{48}\mathbf{}\mathit{m}$

Explanation for the correct answer:

Step 1: Given

Let P be the house $1$ and Q be the house $2$.

Let the height of house P = $100m$ and

The height of the window of the house Q = $64m$

The angle at C and E subtends $90°$. This forms the right angled triangle as shown in the figure.

Step 2: Finding the angles

We know that,

$\tan \left(\theta \right)=\frac{Perpendicular}{Base}$

In $∆BCD$,

$\tan \left(\theta \right)=\frac{64}{d}$ ……….. (1)

Also, in $∆AED,$

$$\begin{gathered} \tan \left(\theta \right)=\frac{d}{100-64} \\ =\frac{d}{36} \end{gathered}$$………… (2)

Step 3: Finding the distance

From equation (1) and (2),

$$\begin{gathered} \Rightarrow \frac{64}{d}=\frac{d}{36} \\ \Rightarrow d^2=64\times 36 \\ \Rightarrow d=\sqrt{64\times 36} \\ \Rightarrow d=8\times 6 \\ \Rightarrow d=48m \end{gathered}$$

Hence, option (A) is the correct answer.