A round balloon of radius `r` subtends an angle at the eye of the observer while the angle of elevation of its centre is `betadot` Prove that the height of the centre of the balloon is `rsinbetacos e calpha/2dot`

Explanation

Let C be the cerntre of the balloon of radius r and 'O' be the point of observation. OA and OB are the tangents from O to the circle.
Join CA, CB and CO.
Draw `CD_|_OX`.
Now, `angle AOB=alpha`
`rArr angleAOC=angle BOC=alpha/2`
(`:.` tangent are equally inclined at the tangent)
`angleDOC=beta`
( `:.` radius through point of contact is `_|_` to the tangent)
First we shall find OC with the help of BC = r and then with the help of OC, we shall find DC (Quite simple)
In right `DeltaOBC`,
`cosecalpha/2=(OC)/(BC)`
`OC=BC. cosec alpha/2=rcosecalpha/2`
In right `DeltaODC`,
`sinbeta=(CD)/(OC)rArrCD=OC sin beta`
`CD =rcosecalpha/2sinbeta=rsinbetacosecalpha/2" [from(1)]"`
Hence, the height of the centre of the balloon from the ground is `r sin beta cosec alpha/2`.