Question

The shadow of the tower standing on a level ground is found to be $60$ meters longer when the suns altitude is ${30}^o$ than when it is ${45}^o$. The height of the tower is

• A. $60m$
• B. $30m$
• C. $60\sqrt{3}m$
• D. $20(\sqrt{3}+1)m$

Answer 1

The correct option is D $20(\sqrt{3}+1)m$

REF.Image

In $△XYZ$

$tan{45}^{\circ }=\frac{XY}{YZ}=\frac{XY}{x}$

$I=\frac{XY}{x}$

$XY=x$ cm

In $△XYA$

$tan{30}^{\circ }=\frac{XY}{AY}$

$\frac{1}{\sqrt{3}}=\frac{x}{60+x}$

$60+x=\sqrt{3}x$

$\frac{60}{(\sqrt{3}-1)}=x$

$x=81.96m$

or Height of Tower = 82 m

or $\frac{60}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=\frac{60(\sqrt{3}+1)}{(\sqrt{3}{)}^2-(1{)}^2}$

$=\frac{60(\sqrt{3}+1)}{3-1}$

$=20(\sqrt{3}+1)$