Question

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is ${60}^{\circ }$. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is ${30}^{\circ }$. Find the height of the tower and the width of the canal.

Answer 1

Here, AB is the height of the tower and BC is the width of canal.
CD = 20 m
As per question,
In right $\Delta ABD,$
tan ${30}^{\circ }=\frac{AB}{BD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{(20+BC)}$
$\Rightarrow AB=\frac{(20+BC)}{\sqrt{3}}...\left(i\right)$
Also,
In right $\Delta ABC,$
tan ${60}^{\circ }=\frac{AB}{BC}$
$\Rightarrow \sqrt{3}=\frac{AB}{BC}$
$\Rightarrow AB=\sqrt{3}BC...\left(ii\right)$
From equation (i) and (ii)
$AB=\sqrt{3}BC=\frac{(20+BC)}{\sqrt{3}}$
$\Rightarrow 3BC=20+BC$
$\Rightarrow 2BC=20\Rightarrow BC=10m$
Putting the value of BC in equation (ii)
$AB=10\sqrt{3}m$
Thus, the height of the tower is $10\sqrt{3}$ m and the width of the canal is 10 m.