Question

Two building are in front of each other on either side of a road of width $10$ metre. From the top of the first building which is $40$ metre high, the angle of elevation to the top of the second is ${45}^{\circ }$. What is the height of the second building?

Answer 1

In the fig., $AB$ and $CD$ represent two building.
$\therefore AB=40m$
$BD$ is the width of the road
$\therefore BD=10m$
$\therefore \angle CAE={45}^{\circ }$
Draw $segAE\parallel segBD$ meeting $segCD$ at point $E$
$\square ABDE$ is a rectangle
$\therefore AE=BD=10m$
and $AB=ED=40m$
Let $CD=x$
In right angle $△AEC$,
$\tan {45}^{\circ }=\frac{CE}{AE}$
$1=\frac{x}{10}$
$\therefore x=10m$
$\therefore CE=10m$
Now, $CD=CE+ED=10+40=50m$
Thus, the height of the second building is $50$ metre.