The length of the shadow of a tower standing on level ground is found to 2x meter longer when the sun’s elevation is 30° than when it was 45 °. The height of the tower in meters is

Answer: Option A

Solution

Answer: Option A

Solution:

AB is a tower
BD and BC are its shadows and CD = 2x

$$\begin{array}{rl}& \tan 45^{\circ }=\frac{AB}{DB}\\ & \Rightarrow 1=\frac{h}{y}\Rightarrow y=h\\ & \text{and}\tan 30^{\circ }=\frac{AB}{CB}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{2x+y}\\ & \Rightarrow 2x+y=\sqrt{3}h\\ & \Rightarrow \sqrt{3}h-h=2x\\ & \Rightarrow h\left(\sqrt{3}-1\right)=2x\\ & \Rightarrow h=\frac{2x}{\sqrt{3}-1}\\ & =\frac{2x\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ & =\frac{2x\left(\sqrt{3}+1\right)}{3-1}\\ & =\frac{2x\left(\sqrt{3}+1\right)}{2}\\ & =x\left(\sqrt{3}+1\right)\end{array}$$