Question

A tower on a hill subtends the same angle at two points A and B on the level ground and the angles of elevation of the top of the tower from these points are respectively $\alpha$ and $\beta$. If the tower and the two points of observation are in the same vertical plane, prove that the height of the tower is
$AB\frac{|cos(\alpha +\beta )|}{sin(\alpha -\beta )}$

Answer 1

Let PQ represent the tower in a hill OP and A, B the points of observation on the ground. Since PQ subtends the same angle at A and B, the four point Q, P, A and B are concyclic. Since angle of elevation of Q at A and B arc respectively $\alpha$ and $\beta$, we have
$\angle QAO=\alpha$ and $\angle QBO=\beta$ so that
$\angle BQA=(\alpha -\beta )$
Let C be the centre of circle.
Now draw CM and CR perpendiculars to the chords AN and PQ so that $AM=BM=\frac{1}{2}AB$
and $PR=QR=\frac{1}{2}PQ$. Join CA, CB, CP and CQ.
Now $\angle ACB=2\angle AQB=2(\alpha -\beta ).$
Hence $\angle ACM=\angle BCM=\alpha -\beta .$
If $\angle PBQ=\theta$; then $\angle PCQ=2\theta$
so that $\angle PCR=\angle QCR=\theta$
We have $\beta ={90}^o-\alpha +\theta$
so that $\theta =\alpha +\beta -{90}^o$
Now $BC=CQ=\left(\frac{x}{2}\right)cosec\theta$ where PQ = x.
Then $BC\sin (\alpha -\beta )=BM=\frac{1}{2}AB$.
or $x=\frac{\sin \theta }{\sin (\alpha -\beta )},AB=\frac{\sin (\alpha +\beta -{90}^o)}{\sin (\alpha -\beta )}$
Hence, $PQ=x=\frac{|\cos (\alpha -\beta )|}{\sin (\alpha -\beta )}AB$.