Question

If the angles of elevation of the top of a tower from two points at distances a and b from the base and in the same straight line with it are complementary then the height of the tower is

(a) $\sqrt{\frac{a}{b}}$

(b) $\sqrt{ab}$

(c) $\sqrt{a+b}$

(d) $\sqrt{a-b}$

Answer 1

$BD=bBC=a\angle ACB=\theta \angle ADC=90-\theta LetAB=htan\theta =\frac{AB}{BC}=\frac{h}{a}----\left(1\right)tan(90-\theta )=\frac{AB}{BD}=\frac{h}{b}----\left(2\right)buttan(90-\theta )=cot\theta \therefore cot\theta =tan\theta =\frac{h}{b}\text{from (1) and (2)}\frac{h}{a}=\frac{h}{b}{h}^2=abh=\sqrt{ab}$

Height of the tower is $\sqrt{ab}$