Question

A flag staff is on the top of tower which stands on a horizontal plane. A person observes the angles, $\alpha$ and $\beta$, subtended at a point on the horizontal plane by the flag staff and the tower; he then walks a known distance a towards the tower and finds that the flag staff subtends the same angle as before; prove that the height of the tower and the length of the flag staff are respectively
$\frac{asin\beta cos(\alpha +\beta )}{cos(\alpha +2\beta )}$ and $\frac{asin\alpha }{cos(\alpha +2\beta )}$

Answer 1

At the point A on the horizontal plane the tower OP = d subtends an angle $\beta$ whereas the flag staff PQ = h subtends an angle $\alpha$. Again PQ subtends the same angle $\alpha$ at a point B such that AB = $\alpha$. Since
$\angle PQA=\angle PQB=\alpha$,
therefore a circle through P and Q will pass through A and B as we know that angle in the same segments are equal. By this property angles subtended by PB at A is $\beta$ so that angle subtends at Q is also $\beta$. Now we shall try to have angle as are required in the solution i.e. $\alpha +2\beta$ and $\alpha +\beta$. $\Delta AOQ$ is a right angles triangle with one angle as $\alpha +\beta$ at A so that other angle at Q is ${90}^o-(\alpha +\beta )=\angle AQP.$
$\therefore \angle AQB={90}^o-(\alpha +\beta )-\beta$
$={90}^o-\left(\alpha 2\beta \right)=\alpha APB$
(Same segment AB)
$\therefore \angle ABP={180}^o-\beta -\angle APB$
$=180-\beta -({90}^o-[\alpha +2\beta \left]\right)$
$={90}^o-(\alpha +\beta )$
We have to find the values of d and h in terms of a and known angles $\alpha$ and $\beta$.
Now $d=APsin\beta$
By sine rule on $\Delta ABP$, we have
$\frac{AB}{sin\left(APB\right)}=\frac{AP}{sin\left(ABP\right)}$
or $\frac{a}{sin({90}^o-(\alpha +2\beta \left)\right)}=\frac{AP}{sin({90}^o-(\alpha -\beta )}$
$\therefore AP=\frac{acos(\alpha +\beta )}{cos\left(\alpha 2\beta \right)}$
Hence from (1) and (2), we have
$d=APsin\beta$
$\frac{acos(\alpha +\beta )sin\beta }{cos(\alpha +2\beta )}=$ height of tower
Now we have to find the height of flag staff PQ. We know the value of AP and hence by sine rule on $\Delta APQ$, we have
$\frac{h}{sinPAQ}=\frac{AP}{sinPQA}$
or $h=AP.\frac{sin\alpha }{sin({90}^o-(\alpha +\beta \left)\right)}=\frac{APsin\alpha }{cos(\alpha +\beta )}$
Putting the value of AP from (2) in the above, we get
or $h=\frac{asin\alpha }{cos(\alpha +2\beta )}$