Question

A vertical tree stands on a hill side that makes an angle $\alpha$ with the horizontal. From a point directly up the hill from the tree, the angle of elevation of three top is $\beta$. From a point m cm, further up the hill, the angle of depression of the tree top is $\gamma$. If the tree is H cm. tall, express H in terms of $\alpha ,\beta ,\gamma$.

Answer 1

In the figure below, OP represents the tree and A and B the two points of observations in the hill inclined at an angle $\alpha$ to the horizon. The elevation of P at A and $\beta$ and the depression of P at B is $\gamma$. We clearly have
$\angle POA={90}^o-\alpha$, $\angle OAP=\alpha +\beta$,
$\angle ABP=\alpha -\gamma$
and $\angle ABP=(\alpha +\beta )-(\alpha -\gamma )=\beta -\gamma$.
Also OP = H cm and AB = m cm.
H comes in $\Delta OPA$ and m comes in $\Delta BPA$ and both have side PA common, which is to be eliminated.
Now from $\Delta OAP$,
$\frac{PA}{sin({90}^o-\alpha )}=\frac{OP}{sin(\alpha +\beta )}$
or $\frac{PA}{cos\alpha }=\frac{H}{sin(\alpha +\beta )}$ ...(1)
and from $\Delta PAB$,
$\frac{AB}{sin(\beta +\gamma )}=\frac{PA}{sin(\alpha -\gamma )}$
or $\frac{m}{sin(\beta +\gamma )}=\frac{PA}{sin(\alpha -\gamma )}$ ...(2)
Multiplying (1) and (2), we get
$\frac{m.PA}{cos\alpha sin(\beta +\gamma )}=\frac{H.PA}{sin(\alpha +\beta )sin(\alpha +\gamma )}$
or $H=\frac{msin(\alpha +\beta )sin(\alpha -\gamma )}{cos\alpha sin(\beta +\gamma )}$