Question

A flag-staff is on the top of a tower standing on a level plane. At certain point in the plane the tower subtends an angle $\alpha$ and the flastaff an angle $\beta$. At another point a ft. nearer the base of the tower, the flag-staff again subtends an angle $\beta$. Prove that the height of the tower is
$\frac{atan\alpha }{1-tan\alpha tan(\alpha +\beta )}$ and the length of the flag-staff
is $\frac{asin\beta }{cos(2\alpha +\beta )}$

Answer 1

$\angle BQC=\angle BPC=\beta$. Hence a circle will pass through B, C, P, Q. Angle subtended by PB and Q and hence at C is also is $\alpha$.
$\therefore \angle ABP=\alpha +\beta$ (sum of interior opposite angles)
We have to find the values of x and y in the given form, where AB = y and BC = x.
We have the following relations:
$y=QAtan\alpha$ ...(1)
$x+y=QAtan(\alpha +\beta )$ ...(2)
$PA=ytan(\alpha +\beta )=(x+y)tan\alpha$ ...(3)
$PA=ytan(\alpha +\beta )=(x+y)tan\alpha$
$QA=QP+PA=a+PA$
or $QA=a+ytan(\alpha +\beta )$, by (3)
$\therefore y=[a+ytan(\alpha +\beta \left)\right]tan\alpha$ by (1)
$\therefore y[1-tan\alpha tan(\alpha +\beta \left)\right]=atan\alpha$
$\therefore y=etc$
$\therefore y=\frac{atan\alpha }{1-tan\alpha tan(\alpha +\beta )}$
Again from (3),
$xtan\alpha =y\left[tan\right(\alpha +\beta )-tan\alpha ]$. Put for y
$\therefore xtan\alpha =\frac{atan\alpha \left[tan\right(\alpha +\beta )-tan\alpha ]}{1-tan\alpha tan(\alpha +\beta )}$
$\therefore x=\frac{asin(\alpha +\beta -\alpha )}{cos(\alpha +\alpha +\beta )}=\frac{asin\beta }{cos(2\alpha +\beta )}$