Question

The shadow of a vertical tower on level ground increases by $20$m, when the altitude of the sun changes from angle of elevation ${60}^o$ to ${45}^o$. Find the height of the tower.

• A. $47.32$m
• B. $47$m
• C. $48$m
• D. $49.56$m

Answer 1

The correct option is A $47.32$m

Let the height of the tower be $h$

In right $\Delta OBC$

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

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

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

In right $\Delta OAC$.

$tan{45}^0=\frac{h}{AB+BC}$

$\Rightarrow tan{45}^0=\frac{h}{20+x}$

$=>h=20+x$ ..............(2)

From (1) & (2) we get

$\sqrt{3x}=20+x$

$\Rightarrow x=\frac{20}{\sqrt{3-1}}$

$\Rightarrow x=\frac{20\sqrt{3+1}}{2}$

$\Rightarrow x=10(\sqrt{3}+1)$

$=27.32$ metres. [since $\sqrt{3}=1.732$]

$\therefore h=20+x$

$=20+27.32$

$=47.32m$

Thus the height of the tower is $47.32m$