Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.

Let PQ be the tower.

Let one of the persons, A be at a distance of x meters and the second person B be at a distance of y metres from the foot of the tower (Q).

Given, that angle of elevation of A is 30°.

From figure,

In △PQA,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{PQ}{AQ} \\[1em] \Rightarrow AQ = PQ\sqrt{3} = 50\sqrt{3} \\[1em] \Rightarrow x = 50 \times 1.732 \\[1em] \Rightarrow x = 86.6 \text{ meters}.$

Given, that angle of elevation of B is 38°.

In △PBQ,

$\Rightarrow \text{tan 38°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7813 = \dfrac{PQ}{BQ} \\[1em] \Rightarrow BQ = \dfrac{50}{0.7813} \\[1em] \Rightarrow y = 64 \text{ meters}.$

AB = x + y = 86.6 + 64 = 150.6 meters.

Hence, the distance between two persons = 150.6 meters.