Question

A round balloon of radius r subtends an angle $\alpha$ at the eye of the observer while the angle of elevation of its centre is $\beta$. Prove that the height of the centre of the balloon is

$\left(rsin\beta cosec\frac{\alpha }{2}\right)$ [3 MARKS]

Answer 1

Let us represent the balloon by a circle with centre C and radius r. Let OX be the horizontal ground and let O be the point of observation. From O, draw tangents OA and OB to the circle. Join CA, CB and CO. Draw $CD\perp OX$

$\therefore \angle AOB=\alpha ,\angle DOC=\beta$ and

$\angle AOC=\angle BOC=\frac{\alpha }{2}$

From right $\Delta OAC$, we have [1 MARK]



$\frac{OC}{AC}=cosec\frac{\alpha }{2}$

$\Rightarrow \frac{OC}{r}=cosec\frac{\alpha }{2}$

$\Rightarrow OC=rcosec\frac{\alpha }{2}$

From right $\Delta ODC$, we have

$\frac{CD}{OC}=sin\beta \Rightarrow CD=\left(OC\right)\times sin\beta$

$\Rightarrow CD=rsin\beta cosec\frac{\alpha }{2}$ [using (i)]

Hence, the height of the centre of the balloon from the ground is $rsin\beta cosec\frac{\alpha }{2}$

Let the height of the centre of the balloon be h

Let the distance between the eye of observer and centre of balloon be L

In $\Delta OBC$ [1 MARK]

$sin\frac{\alpha }{2}=\frac{r}{L}$

$L=\frac{r}{sin\frac{\alpha }{2}}=rcosec\frac{\alpha }{2}\dots \dots \left(1\right)$ [1 MARK]

In $\Delta OCA$

$sin\beta =\frac{h}{L}$

$sin\beta =\frac{h}{rcosec\frac{\alpha }{2}}$

$h=rsin\beta cosec\frac{\alpha }{2}$ [1 MARK]