The angle of elevation of the top of a lighthouse 60 m high, from two points on the ground on its opposite sides are 45° and 60°. What is the distance between these two points?

Answer: Option D

Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
∠BAD = 45°, ∠BCD = 60°
$$\begin{array}{rl}& \tan 45^{\circ }=\frac{BD}{BA}\\ & \Rightarrow 1=\frac{60}{BA}\\ & \Rightarrow BA=60m........\left(\text{i}\right)\\ & \tan 60^{\circ }=\frac{BD}{BC}\\ & \Rightarrow \sqrt{3}=\frac{60}{BC}\\ & \Rightarrow BC=\frac{60}{\sqrt{3}}\\ & =\frac{60\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\ & =\frac{60\sqrt{3}}{3}\\ & =20\sqrt{3}\\ & =20\times 1.73\\ & =34.6m..........\left(\text{ii}\right)\end{array}$$
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m