Question

From the top of a spire, the angles of depression of the top and bottom of a tower of height $h$ are $\theta$ and $\varphi$. show that the height of the spire and its horizontal distance from the tower are respectively
$h.\frac{\cos \theta \sin \varphi }{\sin (\varphi -\theta )}$ and $\frac{h\cos \theta \cos \varphi }{\sin (\varphi -\theta )}$

Answer 1

$y-h=x\tan \theta ,$ from $△BQR.$
$y=x\tan \varphi ,$ from $△BPA$.
$\therefore \frac{y-h}{y}=\frac{\tan \theta }{\tan \varphi }=\frac{\sin \theta \cos \varphi }{\cos \theta \sin \varphi }$
$\therefore 1-\frac{h}{y}=\frac{\sin \theta \cos \varphi }{\cos \theta \sin \varphi }$
or $\frac{\sin \varphi \cos \theta -\cos \varphi \sin \theta }{\sin \varphi \cos \theta }=\frac{h}{y}$
$\therefore y=\frac{h\sin \varphi \cos \theta }{\sin (\varphi -\theta )}$
Again $y=x\tan \varphi$
or $\frac{h\sin \varphi \cos \theta }{\sin (\varphi -\theta )}=x\frac{\sin \varphi }{\cos \varphi }$
$\therefore x=h\frac{\cos \theta \cos \varphi }{\sin (\varphi -\theta )}$
Note : You can prove it by sine rule on triangle $BQR$ and $BQP$ and eliminating $BQ$. This man also be done by $m-n$ theorem on $△BPR$.
$\alpha =\varphi -\theta ,\beta =\theta ,\theta \to 90-\theta$.