The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is

Answer: Option C

Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD = y

$$\begin{array}{rl}& \text{Now in right }\Delta ADB,\\ & \tan \theta =\frac{\text{Perpendicular}}{\text{Base}}=\frac{AB}{DB}\\ & \tan 45^{\circ }=\frac{100}{y}\\ & \Rightarrow 1=\frac{100}{y}\\ & \Rightarrow y=100\\ & \text{Similarly in right }\Delta ACB,\\ & \tan 30^{\circ }=\frac{AB}{CB}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{y+x}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{100+x}\\ & \Rightarrow 100+x=100\sqrt{3}\\ & \Rightarrow x=100\sqrt{3}-100\\ & \Rightarrow x=100\left(\sqrt{3}-1\right)m\end{array}$$