Question

A T.V tower stands vertically on a bank of canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is ${60}^o$. From a point $20m$ away from this point on the same bank, the angle of elevation of the top of the tower is ${30}^o$. Find the height of the tower and the width of the canal.

• A. Height$=16m$; Width$=5m$
• B. Height$=17.32m$; Width$=9m$
• C. Height$=17m$; Width$=9m$
• D. None of these
  

Answer 1

The correct option is D None of these

Let the height of the tower = $h$
Let the width of the tower = $w$
Angle of elevation from the bank of the river = ${60}^{\circ }$
Angle of elevation of a point 20 m far from the bank = ${30}^{\circ }$
The ground, tower and the line of sight forms a right angled triangle.
Thus,
$\tan \angle$ of elevation = $\frac{height}{distanceonground}$
Thus, $\tan 60=\frac{h}{w}$
$w\sqrt{3}=h$ (1)
$\tan 30=\frac{h}{w+20}$
$w+20=h\sqrt{3}$ (2)
From, (1) and (2)
$w\sqrt{3}=\frac{w+20}{\sqrt{3}}$
$3w=w+20$
$w=10$ m
and, $h=w\sqrt{3}=10\sqrt{3}=17.32$ m