An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters higher is the one than the other?

Answer: Option A

Solution

Answer: Option A

Solution:

Let C and D be the position of the aeroplanes.
Given that CB = 900 m, ∠ CAB = 60°, ∠ DAB = 45°
$$\begin{array}{rl}& \text{From}\text{the}\text{right}\Delta ABC,\\ & \tan 60^{\circ }=\frac{CB}{AB}\\ & \sqrt{3}=\frac{900}{AB}\\ & AB=\frac{900}{\sqrt{3}}\\ & =\frac{900\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\ & =\frac{900\sqrt{3}}{3}\\ & =300\sqrt{3}\\ & \text{From}\text{the}\text{right}\Delta ABD,\\ & \tan 45^{\circ }=\frac{DB}{AB}\\ & 1=\frac{DB}{AB}\\ & DB=AB=300\sqrt{3}\\ & \\ & \text{Required}\text{height}\\ & =CD\\ & =\left(CB-DB\right)\\ & =\left(900-300\sqrt{3}\right)\\ & =\left(900-300\times 1.73\right)\\ & =\left(900-519\right)\\ & =381\text{m}\end{array}$$