Question

The width of a road is $b$ feet., on one side of which there is a window $h$ feet height. A building in front of it subtends an angle $\theta$ at it. Prove that the height of the building is $\frac{(b^2+h^2)\sin \theta }{b\cos \theta +h\sin \theta }$

Answer 1

We have to find the value of $BC=x$ in terms of known quantities $b,h$ and $\theta$.
$\therefore h=b\tan \alpha$
$x-h=b\tan (\theta -\alpha )$
$=b\frac{\tan \theta -\tan \alpha }{1+\tan \theta \tan \alpha }$
Put for $\tan \alpha$ from $\left(1\right)$
$\therefore x=h+b\frac{\tan \theta h/b}{1+\tan \theta .h/b}$
$=h+b.\frac{\left(b\sin \theta h\cos \theta \right)}{b\cos \theta +h\sin \theta }$
or $x=\frac{bh\cos \theta +h^2\sin \theta +b^2\sin \theta bh\cos \theta }{b\cos \theta +h\sin \theta }$
$=\frac{(b^2+h^2)\sin \theta }{b\cos \theta +h\sin \theta }$
Note : You can do it by $m-n$ theorem with $\theta ={90}^o,m=ME=h,n=EC-h$.