Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag-staff are $\alpha$ and $\beta$ respectively. Prove that the height of the tower is $\frac{htan\alpha }{tan\beta -tan\alpha }$ [3 MARKS]

Answer 1

Let AB be the tower and BC be the flag-staff. Let O be a point on the plane containing the foot of the tower such that the angles of 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 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

$tan\beta =\frac{y+h}{x}$

$\Rightarrow x=\frac{y+h}{tan\beta }$

$\Rightarrow x=(y+h)cot\beta$ ....... (ii)

On equating the values of x given in equations (i) and (ii), we get

$\Rightarrow ycot\alpha =(y+h)cot\beta$

$\Rightarrow (ycot\alpha -ycot\beta )=hcot\beta$

$\Rightarrow (ycot\alpha -ycot\beta )=hcot\beta$

$\Rightarrow y=\frac{hcot\beta }{cot\alpha -cot\beta }$

$\Rightarrow y=\frac{\frac{h}{tan\beta }}{\frac{1}{tan\alpha -\frac{1}{tan\beta }}}=\frac{htan\alpha }{tan\beta -tan\alpha }$

Hence, the height of the tower is $\frac{htan\alpha }{tan\beta -tan\alpha }$

In $\Delta ABD$

$tan\alpha =\frac{y}{x}$

$x=\frac{y}{tan\alpha }\dots \dots \left(1\right)$ [1 MARK]

In $\Delta ACD$

$tan\beta =\frac{y+h}{x}$

$tan\beta =\frac{y+h}{\frac{y}{tan\alpha }}$ (from (1))

$tan\beta =\frac{(y+h)tan\alpha }{y}$ [1 MARK]

$ytan\beta =ytan\alpha +htan\alpha$

$y(tan\beta -tan\alpha )=htan\alpha$

$y=\frac{htan\alpha }{tan\beta -tan\alpha }$ [$\frac{1}{2}$MARK]