If the angle of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with It are complementary, then the height of the tower is.
A. $a b$ B. $\sqrt{a b}$ C. $\frac{a}{b}$ D. $\sqrt{\frac{a}{b}}$ Answer: Option BShow Answer
Solution(By Apex Team)
Let AB be the tower and P and Q are two points
such that PB = a and QB = b and angles of elevation are
$\theta \text { and }\left(90^{\circ}-\theta\right)$
Let height of tower = h
$\begin{array}{l}\text{Then in right }\triangle\text{APB}\\ \tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{AB}{PB}\\ =\frac{h}{a}\ldots\ldots\ldots\ldots\text{ (i) }\\ \text{Similarily in right }\triangle\text{AQB,}\\ \tan\left(90^{\circ}-\theta\right)=\frac{AB}{QB}=\frac{h}{b}\\ \Rightarrow\cot\theta=\frac{h}{b}\ldots\cdots\cdots\left(ii\right)\\ \text{Multiplying (i) & (ii)}\\ \tan\theta\cot\theta=\frac{h}{a}\times\frac{h}{b}\\ \Rightarrow1=\frac{h^2}{ab}\\ \Rightarrow h^2=ab\\ \Rightarrow h=\sqrt{ab}\\ \therefore\text{height of tower}\\ =\sqrt{ab}\end{array}$
$\begin{array}{l}\text{Then in right }\triangle\text{APB}\\ \tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{AB}{PB}\\ =\frac{h}{a}\ldots\ldots\ldots\ldots\text{ (i) }\\ \text{Similarily in right }\triangle\text{AQB,}\\ \tan\left(90^{\circ}-\theta\right)=\frac{AB}{QB}=\frac{h}{b}\\ \Rightarrow\cot\theta=\frac{h}{b}\ldots\cdots\cdots\left(ii\right)\\ \text{Multiplying (i) & (ii)}\\ \tan\theta\cot\theta=\frac{h}{a}\times\frac{h}{b}\\ \Rightarrow1=\frac{h^2}{ab}\\ \Rightarrow h^2=ab\\ \Rightarrow h=\sqrt{ab}\\ \therefore\text{height of tower}\\ =\sqrt{ab}\end{array}$
