 # 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?

A. $\sqrt{a+b}$ B. $\sqrt{a b}$ C. $\sqrt{a-b}$ D. $\sqrt{\frac{a}{b}}$ 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 $\begin{array}{l}\text{Let AB }=\text{h}\\ \text{Now in right }\triangle\text{APB,}\\ \tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{\text{AB}}{\text{PB}}\\ \Rightarrow\tan30^{\circ}=\frac{\text{h}}{a}\\ \Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{h}}{a}\ldots\ldots\ldots.(i)\\ \text{Similarily in right }\triangle\text{AQB,}\\ \tan60^{\circ}=\frac{\text{AB}}{\text{QB}}\\ \Rightarrow\sqrt{3}=\frac{\text{h}}{\text{b}}\ldots\ldots\ldots\ldots(ii)\\ \text{Multipling (i) & (ii) we get,}\\ \frac{1}{\sqrt{3}}\times\sqrt{3}=\frac{\text{h}}{a}\times\frac{\text{h}}{\text{b}}\\ \Rightarrow1=\frac{\text{h}^2}{\text{ab}}\\ \Rightarrow\text{h}^2=\text{ab}\\ \Rightarrow\text{h}=\sqrt{\text{ab}}\\ \therefore\text{Height of the tower}\\ =\sqrt{\text{ab}}\end{array}$