The angle of elevation of an aeroplane as observed from a point 30 m above the transport water-surface of lake is 30° and the angle of depression of the image of the aeroplane in the water of the lake is 60°. The height of the aeroplane from the water-surface of the lake is

Answer: Option A

Solution

Answer: Option A

Solution:

$$\begin{array}{rl}& \Delta ABE\\ & \tan 30^{\circ }=\frac{AB}{EB}\\ & \frac{1}{\sqrt{3}}=\frac{h}{x}\\ & \Rightarrow x=\sqrt{3}\times h\\ & \Delta EBD\\ & \tan 60^{\circ }=\frac{BD}{EB}\\ & \sqrt{3}=\frac{h+30+30}{x}\\ & \sqrt{3}=\frac{h+60}{\sqrt{3}h}\\ & 3h=h+60\\ & 2h=60\\ & h=30\text{m}\\ & \text{Height from water surface}\\ & =30+30\\ & =60\text{m}\end{array}$$