Question

The angle of elevation of 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

$\tan \theta =\frac{h}{s}$

$\tan (90-\theta )=\frac{h}{t}$

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

$\tan \theta \cdot \cot \theta =\frac{h^2}{st}$

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

$\Rightarrow h=\sqrt{st}$