Two lamp posts AB and CD, each of height 100 m, are on either side of the road. P is a point on the road between the two lamp posts. The angles of elevation of the top of the lamp post from the point P are 60° and 40°. Find the distances PB and PD.

Given,

AB = 100 m

CD = 100 m

∠APB = 40°

∠CPD = 60°

Let PB = x and PD = y.

In triangle ABP,

In triangle ABP,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 40° = \dfrac{AB}{BP} \\[1em] \Rightarrow \tan 40° = \dfrac{100}{x} \\[1em] \Rightarrow 0.8391 = \dfrac{100}{x} \\[1em] \Rightarrow x = \dfrac{100}{0.8391} \\[1em] \Rightarrow x = 119.17 \text{ m.}$

PB = x = 119.17 m

In triangle CDP,

We know that,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60° = \dfrac{CD}{PD} \\[1em] \Rightarrow \sqrt{3} = \dfrac{100}{y} \\[1em] \Rightarrow y = \dfrac{100}{\sqrt{3}} \\[1em] \Rightarrow y = \dfrac{100}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow y = \dfrac{100\sqrt{3}}{3} \text{ m.}$

PD = y = $\dfrac{100\sqrt{3}}{3}$ m

Hence, the distances are PB = 119.17 m and PD = $\dfrac{100\sqrt{3}}{3}$ m.