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

Solution(By Examveda Team)

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

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