If the angles of elevation of a tower from two points distance a and b (a > b) from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is?

Answer: Option B

Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively

$$\begin{array}{rl}& \text{Let }AB=h\\ & \text{Now in right }\Delta APB,\\ & \tan \theta =\frac{\text{Perpendicular}}{\text{Base}}=\frac{AB}{PB}\\ & \Rightarrow \tan 30^{\circ }=\frac{h}{a}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{a}...........(i)\\ & \text{Similarly in right }\Delta AQB,\\ & \tan 60^{\circ }=\frac{AB}{QB}\\ & \Rightarrow \sqrt{3}=\frac{h}{b}...........(ii)\\ & \text{Multiplying (i) and (ii)}\\ & \frac{1}{\sqrt{3}}\times \sqrt{3}=\frac{h}{a}\times \frac{h}{b}\\ & \Rightarrow 1=\frac{h^2}{ab}\\ & \Rightarrow h^2=ab\\ & \Rightarrow h=\sqrt{ab}\\ & \therefore \text{Height of the tower}=\sqrt{ab}\end{array}$$