Question

A pole stands in a park such that its shadow increases by 2 m when the angle of elevation of Sun changes from 45º to 30º. The height of the pole is

• A. $(\sqrt{3}-1)m$
• B. $2\sqrt{3}m$
• C. $(\sqrt{3}+1)m$
• D. $\sqrt{3}m$

Answer 1

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

Let AO be the pole of height h.
$In\Delta AOB,$
$\tan {45}^{\circ }=\frac{AO}{OB}$
$\Rightarrow 1=\frac{h}{x}$
$\Rightarrow h=x$ ....(i)
$In\Delta AOC,$
$\tan {30}^{\circ }=\frac{AO}{OC}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+2}$
$\Rightarrow x+2=\sqrt{3}h$
$\Rightarrow \sqrt{3}h-x=2$
$\Rightarrow (\sqrt{3}-1)h=2$ [From (i)]
$\Rightarrow h=\frac{2}{\sqrt{3}-1}$
$\Rightarrow h=\frac{2}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\Rightarrow h=\frac{2}{2}(\sqrt{3}+1)$
$\Rightarrow h=(\sqrt{3}+1)m$
Thus, the height of the pole is $(\sqrt{3}+1)m$.
Hence, the correct answer is option (c).