Question

If the angles of elevation of the top of a tower form tow 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

(b) $\sqrt{ab}$
Let $AB$ be the tower and $C$and $D$ be the points of observation on $AC$.
Let:
∠$ACB=\theta$, ∠$ADB=90-\theta$ and $AB=h$ m
Thus, we have:
$AC=a,AD=b$ and $CD=a-b$

Now, in the right ∆ABC, we have:
$\tan \theta =\frac{AB}{AC}$ ⇒ $\frac{h}{a}=\tan \theta$ ...(i)
In the right ∆ABD, we have:
$\tan (90-\theta )=\frac{AB}{AD}$ ⇒ $cot\theta =\frac{h}{b}$ ...(ii)

On multiplying (i) and (ii), we have:
$\tan \theta \times cot\theta =\frac{h}{a}\times \frac{h}{b}$
⇒ $\frac{h}{a}\times \frac{h}{b}=1$ [ ∵ $\tan \theta =\frac{1}{cot\theta }$]
⇒ $h^2=ab$
⇒ $h=\sqrt{ab}$ m

Hence, the height of the tower is $\sqrt{ab}$ m.