Question

The elevation of a tower due North of a station $A$ is $\alpha$ or (at a place due South of it is $\alpha$) and at another station $B$ due West of $A$ it is $\beta$. Prove that the height of the tower is

$\frac{AB}{\sqrt{({\cot }^2\beta -{\cot }^2\alpha )}}$

or $\frac{AB\tan \alpha \tan \beta }{\sqrt{({\tan }^2\alpha -{\tan }^2\beta )}}$

or $\frac{AB\sin \alpha \sin \beta }{\sqrt{({\sin }^2\alpha -{\sin }^2\beta )}}$

or $\frac{AB\sin \alpha \sin \beta }{{\left[\sin (\alpha +\beta )\sin (\alpha -\beta )\right]}^{1/2}}$ if ${\sin }^2\alpha -{\sin }^2\beta =\sin (\alpha +\beta )\sin (\alpha -\beta )$

Answer 1

$h=PA\tan \alpha$

$\therefore PA=h\cot \alpha$

and $h=PB\tan \beta$

$\therefore PB=h\cot \beta$

Also from the right-angled triangle $PAB$, we get

${PB}^2={PA}^2+{AB}^2$

$\therefore {AB}^2={PB}^2-{PA}^2$

$=h^2({\cot }^2\beta -{\cot }^2\alpha )$

$\therefore h=\frac{AB}{\sqrt{({\cot }^2\beta -{\cot }^2\alpha )}}$

$h=\frac{AB\tan \alpha \tan \beta }{\sqrt{{\tan }^2\alpha -{\tan }^2\beta }}$

Substituitng, $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$

Or ,

$h=\frac{AB\sin \alpha \sin \beta }{\sqrt{({\sin }^2\alpha {\cos }^2\beta -{\cos }^2\alpha {\sin }^2\beta )}}$

Substituting $cos\beta =(1-si{n}^2\beta )$ and $\cos \alpha =(1-si{n}^2\alpha )$

$=\frac{AB\sin \alpha \sin \beta }{\sqrt{[{\sin }^2\alpha (1-{\sin }^2\beta )-{\sin }^2\beta (1-{\sin }^2\alpha )]}}$

$=\frac{AB\sin \alpha \sin \beta }{\sqrt{({\sin }^2\alpha -si{n}^2\beta )}}$

$=\frac{AB\sin \alpha \sin \beta }{\sqrt{\left[\sin (\alpha +\beta )\sin (\alpha -\beta )\right]}}$

$\because {\sin }^2\alpha -{\sin }^2\beta =\sin (\alpha +\beta )\sin (\alpha -\beta )$