Question

Question 6
The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is $\sqrt{st}$

Answer 1

Let the height of the tower be h.

And, $\angle ABC=\theta$

Given that, BC = s, PC = t

And, the angle of elevation on both position are complementary.

i.e., $\angle APC={90}^{\circ }-\theta$



[ if two angles are complementary to each other, then the sum of both angles is equal to ${90}^{\circ }$].

In $\Delta ABC$, $tan\theta =\frac{AC}{BC}=\frac{h}{s}\dots \dots \left(i\right)$

And, in $\Delta APC$,

$tan({90}^{\circ }-\theta )=\frac{AC}{PC}$ $[\because tan({90}^{\circ }-\theta )=cot\theta ]$

$\Rightarrow cot\theta =\frac{h}{t}$

$\Rightarrow \frac{1}{tan\theta }=\frac{h}{t}$ $[\because cot\theta =\frac{1}{tan\theta }]\dots \dots \left(ii\right)$

On multiplying Eq.s(i) and (ii), we get;

$tan\theta .\frac{1}{tan\theta }=\frac{h}{s}.\frac{h}{t}$

$\Rightarrow \frac{h^2}{st}=1$

$\Rightarrow h^2=st$

$\Rightarrow h=\sqrt{st}$

So, the required height of the tower is $\sqrt{st}$.