Question

A tower has flag staff at its top which subtends equal angles $\alpha$ at a points distance 9 yds and 11 yds from the foot of the tower. If $tan\alpha =\frac{1}{10}$, find the height of the tower and the flag staff.

Answer 1

AB = y, tower, BC = x, flagstaff, AP = 9, AQ = 11. BC subtends equal angles $\alpha$ at P and Q s.t. $tan\alpha =\frac{1}{10}$. Hence the circle through C and B must also pass through P and Q as angles in the same segment are equal. From the figure it is clear that segment BP subtends equal angles say $\beta$ at C and Q.
$y+11tan\beta$ ...(1)
and $x+y=11tan(\alpha +\beta )$ ...(2)
$AP=9=(x+y)tan\beta$ ...(3)
$\therefore 9=11tan(\alpha +\beta )tan\beta$,
from (2) and (3),
or $9=11tan\beta \left(\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }\right)$
Put $tan\alpha =\frac{1}{10}$.
$\therefore 9[1-(-\frac{1}{10})tan\beta ]=11tan\beta (\frac{1}{10}+tan\beta )$
$9(10-tan\beta )=11tan\beta (1+10tan\beta )$.
$110ta{n}^2\beta +20tan\beta -90=0$
or $11ta{n}^2\beta =2tan\beta -9=0$
$(11tan\beta -9)(tan\beta +1)=0$
$\therefore tan\beta =\frac{9}{11}\left(\beta acute\right)$
Hence from (1),
$y=11tan\beta =11.\frac{9}{11}=9$
From (3),
$9=(x+y)tan\beta =(x+9).\frac{9}{11}$.
$\therefore x+9=11$ or $x=2$.