Question

A vertices tower stands on a horizontal plan and is surrounding by a vertices flag staff of a height h . At a point on the plane , the angle of elevation pf the bottom and the top of the flag staff are the $\alpha and\beta$ respectively. Prove that the height of the tower is $\frac{h\tan \alpha }{\tan \beta -\tan \alpha }$

Answer 1

Let AB be the tower and BC be the flag-staff. Let O be the point on the plane containing the foot of the tower such that the angles at elevation of the bottom B and top C of the flag-staff at O are $\alpha and\beta$ respectively. Let OA = x metres, AB = y metres and BC = h metres.
IN $\Delta OAB$, we have
$tan\alpha =\frac{AB}{OA}$
$\Rightarrow tan\alpha =\frac{y}{x}$
$\Rightarrow x=\frac{y}{tan\alpha }$ .....(i)
$\Rightarrow x=ycot\alpha$
In $\Delta OAC$, we have
$\Rightarrow tan\beta =\frac{y+h}{x}$
$\Rightarrow x=\frac{y+h}{tan\beta }$
$\Rightarrow x=(y+h)cot\beta$ ...(ii)
On equatting the values of x given in equations (i) and (ii), we get
$\Rightarrow ycot\alpha =(y+h)cot\beta$
$\Rightarrow (ycot\alpha -ycot\beta )=hcot\beta$
$\Rightarrow y(cot\alpha -cot\beta )=hcot\beta$
$\Rightarrow y=\frac{hcot\beta }{cot\alpha -cot\beta }$
$\Rightarrow y=\frac{\frac{h}{tan\beta }}{\frac{1}{tan\alpha }-\frac{1}{tan\beta }}=\frac{htan\alpha }{tan\beta -tan\alpha }$
Hence, the height of the tower is $\frac{htan\alpha }{tan\beta -tan\alpha }$