Question

A person walking along a straight road observes that at two points $1km.$ apart, the angles of elevation of a pole in front of him are ${30}^{\circ }$ and ${75}^{\circ }.$ The height of pole is

• A. $250(\sqrt{3}+1)m$
• B. $250(\sqrt{3}-1)m$
• C. $500$ $m$
• D. $500(\sqrt{2}-1)m$

Answer 1

Answer: (A)

$250(\sqrt{3}+1)m$

$\begin{array}{c}In\Delta POQ\\ \tan {75}^{\circ }=\frac{h}{x}here,x=\frac{h}{\tan {75}^{\circ }}\\ In\Delta PAQ\\ \tan {30}^{\circ }=\frac{h}{x+1}\\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+1}\\ \Rightarrow \sqrt{3}h=x+1\\ \Rightarrow \sqrt{3}h=\frac{h}{\tan {75}^{\circ }}+1\\ \Rightarrow \sqrt{3}h-\frac{h}{\tan {75}^{\circ }}=1\\ \Rightarrow h\left[\frac{\sqrt{3}\tan {75}^{\circ }-1}{\tan {75}^{\circ }}\right]=1\\ \Rightarrow h\left[\frac{4}{\sqrt{3}+1}\right]=1\left[onputtingthevsalueof\tan {75}^{\circ }\right]\\ \therefore h=\frac{\sqrt{3}+1}{4}km\\ =250\left(\sqrt{3}+1\right)m\\ Hence,Aistherequiredanswer.\end{array}$