Question

A flag staff on the top of a tower is observed to subtend the same angle $\alpha$ at two points on a horizontal plane, which lie on a line passing through the centre of the base of the tower and whose distance from one another is 2a and an angle $\beta$ at a point half-way between them. Prove that the height of the flag staff is
$asin\alpha \sqrt{\left(\frac{2sin\beta }{cos\alpha sin(\beta -\alpha )}\right)}$

Answer 1

Let OP represent the tower and PQ = 2h the flag staff. Let R, S and M be the points of observation where RS = 2a, so that
$\angle QRP=\alpha =\angle QSP$,
$\angle QMP=\beta$ and $RM=a=MS$.
Since $\angle QRP=\angle QSP$, a circle will pass through the four points P, Q, R and S. Let C be the centre of this circle, and from C draw CL and CM perpendiculars to PQ and RS, bisecting them in L and M respectively so that LP = LQ = h. Join CQ, CP and CR.
Let he $\angle PMO=\varphi$. We then have
Angle at the centre = Twice the angle at the circumference
$\angle PCQ=2\alpha ,$ so that
$\angle PCL=\alpha =\angle LCQ$
Let OL = CM = y, then, $OQ=y+h,OP=y-h$
If $OM=x=CI=hcot\alpha$ [In $\Delta CLQ$]
$\therefore x=hcot\alpha$ ...(1)
Also $CQ=hcosec\alpha =CR$ ...(2)
[From $\Delta CLQ$]
Now from rt. angled $\Delta CRM$
$C{R}^2=R{M}^2+C{M}^2$
or $h^{2}cose{c}^2\alpha -{\alpha }^2=y^2$ ...(3)
We have to find the height PQ = 2h of the flag-staff in terms of known quantities $a,\alpha$ and $\beta$ by eliminating unknown quantities x and y.
Now $tan(\beta +\varphi )=\frac{OQ}{OM}=\frac{y+h}{x}$
and $tan\varphi =\frac{OP}{OM}=\frac{y-h}{x}$
$\therefore \beta =(\beta +\varphi )-\varphi =ta{n}^{-1}\frac{y+h}{x}ta{n}^{-1}\frac{y-h}{x}$
or $\beta =ta{n}^{-1}\frac{\frac{y+h}{x}-\frac{y-h}{x}}{1+\frac{y^2-h^2}{x^2}}$
or $tan\beta =\frac{2hx}{x^2+y^2-h^2}$
Now put the values of x and y from (1) and (3).
$\therefore tan\beta (h^{2}co{t}^2\alpha +h^{2}cose{c}^2\alpha -{\alpha }^2-h^2)$
$=2h^2\left(cot\alpha \right)$
or $tan\beta (2h^{2}co{t}^2\alpha -{\alpha }^2)=2h^{2}cot\alpha$
or ${\alpha }^{2}tan\beta =h^2(2tan\beta co{t}^2\alpha -2cot\alpha )$
$=2h^{2}cot\alpha (tan\beta cot\alpha -1)$
$\therefore a^2\frac{sin\beta }{cos\beta }=2h^2\frac{cos\alpha sin(\beta -\alpha )}{sin\alpha cos\beta sin\alpha }$
or $2h=asin\alpha \sqrt{\left(\frac{2sin\beta }{cos\alpha sin(\beta -\alpha )}\right)}$