Question

The angles of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at distance $a$ and $b$ respectively are complementary angles. Prove that the height of the tower is $\sqrt{ab}$. If the line joining the two points subtends and angle $\theta$ at the top of the tower, show that $\sin \theta =\frac{(a-b)}{(a+b)}$.

Answer 1

$h=a\tan \alpha ,h=b\tan ({90}^o-\alpha )=b\cot \alpha .$
Multiply $h^2=ab\tan \alpha \cot \alpha =ab.$
$\therefore h=\sqrt{ab}$
Also $\tan \alpha =\frac{h}{\alpha }=\frac{\sqrt{ab}}{a}=\sqrt{\left(\frac{b}{a}\right)}$ ...$\left(1\right)$
If $AB$ subtends an angle $\theta$ at $Q$ then
${90}^o-\alpha =\theta +\alpha$
or $\theta ={90}^o-2\alpha$
$\therefore \sin \theta =\cos 2\alpha =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$
or $\sin \theta =\frac{1-b/a}{1+b/a}=\frac{a-b}{a+b}$, by $\left(1\right)$