Question

In the figure given, from the top of a building AB$=60m$ high the angles of depression of the top and bottom of a vertical lamp post $CD$ are observed to be ${30}^{\circ }$ and ${60}^{\circ }$ respectively. Find: (i) The horizontal distance between $AB$ and $CD$. (ii) The height of the lamp post $CD$.

Answer 1

Given: $AB=60m$
$\because \angle PAC={60}^{\circ }$
$\therefore \angle PAC=\angle BCA$
(i) Now in $\Delta ABC$,
$\tan {60}^{\circ }=\frac{AB}{BC}$
$\sqrt{3}=\frac{60}{BC}$
$\Rightarrow BC=\frac{60}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$BC=\frac{60\sqrt{3}}{3}=20\sqrt{3}$
Hence, the horizontal distance between $AB$ and $CD=20\sqrt{3}m$.
(ii) Let $AE=x$ and proved above $BC=20\sqrt{3}m$
$\therefore BC=ED=20\sqrt{3}$
Now in $\Delta AED$,
$\tan {30}^{\circ }=\frac{AE}{ED}$
$\frac{1}{\sqrt{3}}=\frac{AE}{20\sqrt{3}}$
$\Rightarrow \sqrt{3}AE=20\sqrt{3}$
$\Rightarrow AE=20m$
now $EB=AB-AE$
$\therefore EB=60-20=40m$
$\therefore EB=CD$
$\therefore CD=40m$
Hence, the height of the lamp post $=40m$.