Question

The angles of elevation of the top of a vertical tower from two points 30 metres apart and on the same straight line passing through the base of tower are ${30}^{\circ }$ and ${60}^{\circ }$ respectively. The height of the tower is

• A. $10m$
• B. $15m$
• C. $15\sqrt{3}m$
• D. $30m$

Answer 1

Answer: (C)

$15\sqrt{3}m$

Let the height of the tower is h

In $\Delta ABC$

$\tan {60}^0=\frac{h}{x}$

$=>\sqrt{3}=\frac{h}{x}$

$=>x=\frac{h}{\sqrt{3}}$4 ......(i)

In $\Delta ABD$

$\tan {30}^0=\frac{h}{x+30}$

$=>\frac{1}{\sqrt{3}}=\frac{h}{x+30}$

$=>x+30=h\sqrt{3}$ ........(ii)

$=>x=h\sqrt{3}-30$

From (i) and (ii)

$\frac{h}{\sqrt{3}}=h\sqrt{3}-30$

$=>h=3h-30\sqrt{3}$

$=>2h=30\sqrt{3}$

$=>h=15\sqrt{3}$

So, the height of tower = $15\sqrt{3}$m