Question

The angle of elevation of the top of a pillar at any point $A$ on the ground is ${15}^{\circ }$ . On walking $100$ ft. towards the pillar, the angle becomes ${30}^{\circ }$. Height of the pillar and its distance from $A$ are _______ and _______ respectively.

Answer 1

Let $CD$ be the height of the pillar and $AB=100$ ft be the distance travelled by the man.

Let initial angle of elevation $\angle DAC={15}^{\circ }$ and final angle be $\angle DBC={30}^{\circ }$.

Consider, right $\Delta DAC$,

$\Rightarrow \tan {15}^{\circ }=\frac{DC}{AC}=\frac{DC}{100+x}$

$\Rightarrow 2-\sqrt{3}=\frac{DC}{100+x}$ [$\because \tan {15}^{\circ }=2-\sqrt{3}$]

$\Rightarrow (2-\sqrt{3})(100+x)=DC$...(i)

Consider, right $\Delta DBC$,

$\Rightarrow \tan {30}^{\circ }=\frac{DC}{BC}=\frac{DC}{x}$ ...(ii)

From (i), we get,

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{(2-\sqrt{3})(100+x)}{x}$

$\Rightarrow x=\sqrt{3}(200+2x-100\sqrt{3}-\sqrt{3}x)$

$\Rightarrow x=200\sqrt{3}+2\sqrt{3}x-300-3x$

$\Rightarrow x-2\sqrt{3}x+3x=200\sqrt{3}-300$

$\Rightarrow 2\sqrt{3}x(2-\sqrt{3})=100\sqrt{3}(2-\sqrt{3})$

$\Rightarrow x=50$ ft.

Therefore, from (ii)

$\Rightarrow DC=\frac{1}{\sqrt{3}}\times 50=\frac{50\sqrt{3}}{3}$ ft.

Height of the pillar and its distance from $A$ are $\frac{50\sqrt{3}}{3}$ ft and $100+50=150$ ft respectively.