Question

Question 8
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height "h" units. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are $\alpha$ and $\beta$ respectively. Prove that the height of the tower is $\left(\frac{htan\alpha }{tan\beta -tan\alpha }\right)$.

Answer 1

Let the height of the tower be H and OR = x

Given that, height of flag staff = h = FP and $\angle PRO=\alpha ,\angle FRO=\beta$

Now, in $\Delta PRO$, $tan\alpha =\frac{PO}{RO}=\frac{H}{x}$

$\Rightarrow x=\frac{H}{tan\alpha }$ ....(i)

And in $\Delta FRO$, $tan\beta =\frac{FO}{RO}=\frac{FP+PO}{RO}tan\beta =\frac{h+H}{x}$

$\Rightarrow x=\frac{h+H}{tan\beta }\dots \dots \left(ii\right)$

From Eqs.(i) and (ii),

$\frac{H}{tan\alpha }=\frac{h+H}{tan\beta }$

$\Rightarrow Htan\beta =htan\alpha +Htan\alpha$

$\Rightarrow Htan\beta -Htan\alpha =htan\alpha$

$\Rightarrow H(tan\beta -tan\alpha )=htan\alpha \Rightarrow H=\frac{htan\alpha }{tan\beta -tan\alpha }$

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