Question

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

Answer 1

Answer:

48 m.

Step-by-step explanation:

We are given that height of house =100m

Height of windows=64 m

We have to find the distance between the two houses

In triangle DAB

Let AB=d

AB=DE=d

$tan\theta=\frac{64}{d}$

$d=64 cot\theta$

BC=100 m

CE= BC-BE=100-64=36

In triangle CED

$tan(90-\theta )=\frac{36}{d}$

$cot\theta=\frac{36}{d}$

$d=36 tan\theta$ ($tan\theta=\frac{1}{cot\theta})$

$d\times d=36\times 64 cot\theta\times tan\theta==2304$

$d=\sqrt{2304}= 48 m$

Hence, the distance between two houses is 48 m.