A poster is on top of a building. Rajesh is standing on the ground at a distance of 50 m from the building. The angles of elevation to the top of the poster and bottom of the poster are 45° and 30° respectively. What is the height of the poster?

Answer: Option A

Solution

Answer: Option A

Solution:

$$\begin{array}{rl}& \text{in}\Delta PNQ,\tan 45^{\circ }=\frac{PQ}{NQ}\\ & \therefore PQ=NQ=50\text{m}\\ & \text{in}\Delta MNQ,\tan 30^{\circ }=\frac{MQ}{NQ}\\ & \therefore \frac{1}{\sqrt{3}}=\frac{MQ}{50}\\ & \therefore MQ=\frac{50}{\sqrt{3}}\\ & \therefore h=PM\\ & =PQ-MQ\\ & =50-\frac{50}{\sqrt{3}}\\ & =\frac{50}{\sqrt{3}}\left(\sqrt{3}-1\right)\end{array}$$
∴ Poster height = $$\frac{50}{\sqrt{3}}\left(\sqrt{3}-1\right)$$