From the top of a tower, the angles of depression of two objects P and Q (situated on the ground on the same side of the tower) separated at a distance of 100$${\left( {3 - \sqrt 3 } \right)}$$   m are 45° and 60 ° respectively. The height of the tower is-

Solution(By Examveda Team)

$$\eqalign{ & {\text{Let, }}OP = {\text{ }}a \cr & {\text{tan }}{60^ \circ } = \frac{H}{a} \cr & \Rightarrow H = \sqrt 3 a \cr & \Rightarrow \frac{H}{{\sqrt 3 }} = a.....(i) \cr} $$
$$tan{45^ \circ } = 1$$   $$ = \frac{H}{{a + 100\left( {3 - \sqrt 3 } \right)}}$$
$$ \Rightarrow a + 100\left( {3 - \sqrt 3 } \right) = H$$
From (i) $$\frac{H}{{\sqrt 3 }} + $$   $$100\left( {3 - \sqrt 3 } \right)$$   = H
$$\eqalign{ & \Rightarrow H + 300\sqrt 3 - 300 = \sqrt 3 H \cr & \Rightarrow 300\sqrt 3 - 300 = \sqrt 3 H - H \cr & \Rightarrow \left( {\sqrt 3 - 1} \right)H = 300\left( {\sqrt 3 - 1} \right) \cr & \Rightarrow H = 300{\text{ m}} \cr} $$