Question

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30^o than when it is 60^o. Find the height of the tower.

• A. 40√3
• B. 20√3
• C. 20
• D. 40

Answer 1

The correct option is $D$

In tringle ABD,

tan ${60}^0$ $=\frac{AB}{BD}$

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

$\therefore$ $x=\frac{h}{\sqrt{3}}$ ......(1)

Now in triangle ABC,

tan ${30}^0=\frac{AB}{BC}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}$ $=\frac{h}{x+40}$

$\Rightarrow x+40=\sqrt{3h}$

$\therefore$ $\frac{h}{\sqrt{3}}$ $+40=\sqrt{3h}$ [from equ..(1)]

$\Rightarrow h+40\sqrt{3}=3h$

$\Rightarrow 2h=40\sqrt{3}$

$\therefore$ $h=20\sqrt{3}$

∴ The height of a tower is $20\sqrt{3}$