Question

The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line are complemetary. Prove that the height of the tower is $\sqrt{ab}$ metres.

Answer 1

Let AB be the tower and let C and D be the two positions of the observer. Then,



AC = a metres and AD = b metres

Let $\angle ACB=\theta$. Then, $\angle ADB=({90}^{\circ }-\theta )$

Let AB = h metres

From right $\Delta DAB$, we have

$\frac{AB}{AD}=tan({90}^{\circ }-\theta )\Rightarrow \frac{h}{b}=cot\theta$

$\Rightarrow h=bcot\theta$ ............ (i)

From right $\Delta CAB$, we have

$\frac{AB}{AC}=tan\theta \Rightarrow \frac{h}{a}=tan\theta \Rightarrow h=atan\theta$ ............. (ii)

From (i) and (ii), we get $h^2=ab\Rightarrow h=\sqrt{ab}$.

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