Question

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

Answer 1

Step 1: Making the diagram.

Let us consider AC and BD as the two poles of the same height $hm$.

Given $AB=80m$

Let $AP=xm$

Hence, $PB=(80–x)m$

Step 2: Finding the relations.

In $△APC,$

We know that,

$\begin{array}{rcl}& \Rightarrow & \tan \theta =\frac{perpendicular}{Base}\\ & \Rightarrow & \tan \theta =\frac{AC}{AP}\\ & \Rightarrow & \tan 30°=\frac{h}{x}\\ & \Rightarrow & \frac{1}{\sqrt{3}}=\frac{h}{x}\\ & \Rightarrow & x=\sqrt{3}h\end{array}$

Now, $In△BPD,$

$\begin{array}{rcl}& \Rightarrow & \tan \theta =\frac{perpendicular}{Base}\\ & \Rightarrow & \tan \theta =\frac{BD}{PB}\\ & \Rightarrow & \tan 60°=\frac{h}{80-x}\\ & \Rightarrow & \sqrt{3}=\frac{h}{80-x}\\ & \Rightarrow & 80\sqrt{3}-\sqrt{3}x=h\\ & \Rightarrow & 80\sqrt{3}-\sqrt{3}\left(\sqrt{3}h\right)=h\\ & \Rightarrow & 80\sqrt{3}=h+3h\\ & \Rightarrow & h=\frac{80\sqrt{3}}{4}\\ & \Rightarrow & h=20\sqrt{3}\end{array}$

Now, $x=\sqrt{3}h=\sqrt{3}\times 20\sqrt{3}=20\times 3=60m$

Hence, the height of the poles is $20\sqrt{3}m$ and the distance from the pole AC is $60m$ and from the pole BD is $20m$.