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