Question

From the top of $60m$ high building, the angle of elevation and angle of depression of the top and bottom of a light house from top of building are ${30}^0$ and ${60}^0$ respectively. Find out
(i) Difference in heights of light house and building.
(ii) Distance between light house and building.

Answer 1

Let $AB$ is a building of height $60m$ and $CD$ is a light house of height $h$ m. The angle of elevation and angle of depression of the top and bottom of a light house from top of building are ${30}^0$ and ${60}^0$ respectively.
$\angle CAE={30}^0$ and $\angle EAD={60}^0$
$\angle ADB=\angle EAD={60}^0$ (Alternate angle)
Draw $BD||AE$
$\therefore \angle AEC={90}^0$ (Corresponding angle)
$\angle ABD+\angle BDE={90}^0+{90}^0={180}^0$
$\therefore AB||DE$
So, $ABDE$ is a rectangle.
$DE=AB=60m$
and $CE=(h–60)m$
From right angled $\Delta AEC$,
$\tan {30}^0=\frac{CE}{AE}$
$\frac{1}{\sqrt{3}}=\frac{h-60}{BD}[\because AE=BD]$
$BD=\sqrt{3}(h–60)m$ …..(i)
From right angled $\Delta ABD$,
$\tan {60}^0=\frac{AB}{BD}$
$\sqrt{3}=\frac{60}{BD}$
$BD=\frac{60}{\sqrt{3}}=\frac{60\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$
$=20\sqrt{3}$.............(ii)
Put the value in equation (ii) from equation. (i),
$\sqrt{3}(h–60)=20\sqrt{3}$
$h–60=\frac{20\sqrt{3}}{\sqrt{3}}=20$
$h=20+60=80m$
Hence, height of light house $=80m$
(i) Difference in height between light house and building
$=80–60=20m$
(ii) Distance between light house and building
$BD=20\sqrt{3}m$