Question

Two poles of equal height are standing opposite each other on either side of the road which is $80m$ wide.From a point between then on the road the angles of elevation of top of the poles are ${60}^o$ and ${30}^o$. Find the height of the poles and distance of the point from the poles.

Answer 1

Let AB and CD be two poles.

Therefore,

$AB=CD$

Length of road $=80m\left(\text{Given}\right)$

$\therefore BC=80m$

Let point $P$ be the point on the road between poles.

Since poles are perpendicular to the ground.

$\angle ABP=\angle DCP=90°$

As we know that,

$\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

Now, in right angled triangle $DPC$,

$\tan 30°=\frac{CD}{CP}$

$\Rightarrow CD=\frac{CP}{\sqrt{3}}.....\left(1\right)$

Similarly in right angled triangle $APB$,

$\tan 60°=\frac{AB}{BP}$

$\Rightarrow AB=\sqrt{3}BP$

$\Rightarrow CD=\sqrt{3}BP.....\left(2\right)(\because AB=CD)$

From equation $\left(1\right)\&\left(2\right)$, we have

$\frac{CP}{\sqrt{3}}=\sqrt{3}BP$

$\Rightarrow CP=3BP.....\left(3\right)$

Now,

$BC=BP+CP$

$\Rightarrow 80=BP+3BP\left(\text{From}\left(3\right)\right)$

$\Rightarrow BP=\frac{80}{4}=20m$

Now from equation $\left(3\right)$, we have

$CP=3\times 20=60m$

Again from equation $\left(1\right)$, we have

$CD=\frac{60}{\sqrt{3}}=20\sqrt{3}m$

Therefore,

Height of the pole is $20\sqrt{3}m$.

Distance of point $P$ from pole $CD$ is $60m$.

Distance of point from pole $AB$ is $20m$.