Question

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on
the road, the angle of elevation of the top of one pole is 60° and the angle of depression from the top of another pole at P is 30°. Find the
height of each pole and distances of the point P from the poles. [CBSE 2015]

Answer 1

Let AB and CD be the equal poles; and BD be the width of the road.

We have,
$$\begin{gathered} \angle \mathrm{AOB}=60°\mathrm{and}\angle \mathrm{COD}=60° \\ \mathrm{In}∆\mathrm{AOB}, \\ \tan 60°=\frac{\mathrm{AB}}{\mathrm{BO}} \\ \Rightarrow \sqrt{3}=\frac{\mathrm{AB}}{\mathrm{BO}} \\ \Rightarrow \mathrm{BO}=\frac{\mathrm{AB}}{\sqrt{3}} \\ \mathrm{Also},\mathrm{in}∆\mathrm{COD}, \\ \tan 30°=\frac{\mathrm{CD}}{\mathrm{DO}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{CD}}{\mathrm{DO}} \\ \Rightarrow \mathrm{DO}=\sqrt{3}\mathrm{CD} \\ \mathrm{As},\mathrm{BD}=80 \\ \Rightarrow \mathrm{BO}+\mathrm{DO}=80 \\ \Rightarrow \frac{\mathrm{AB}}{\sqrt{3}}+\sqrt{3}\mathrm{CD}=80 \\ \Rightarrow \frac{\mathrm{AB}}{\sqrt{3}}+\sqrt{3}\mathrm{AB}=80\left(\mathrm{Given}:\mathrm{AB}=\mathrm{CD}\right) \\ \Rightarrow \mathrm{AB}\left(\frac{1}{\sqrt{3}}+\sqrt{3}\right)=80 \\ \Rightarrow \mathrm{AB}\left(\frac{1+3}{\sqrt{3}}\right)=80 \\ \Rightarrow \mathrm{AB}\left(\frac{4}{\sqrt{3}}\right)=80 \\ \Rightarrow \mathrm{AB}=\frac{80\sqrt{3}}{4} \\ \Rightarrow \mathrm{AB}=20\sqrt{3}\mathrm{m} \\ \mathrm{Also},\mathrm{BO}=\frac{\mathrm{AB}}{\sqrt{3}}=\frac{20\sqrt{3}}{\sqrt{3}}=20\mathrm{m} \\ \mathrm{So},\mathrm{DO}=80-20=60\mathrm{m} \end{gathered}$$

Hence, the height of each pole is 20$\sqrt{3}$ m and point P is at a distance of 20 m from left pole and 60 m from right pole.