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 {O} be the centre of the balloon and P be the eye of the observer And angle {APB} be the angle subtended by the balloon


therefore `angle {APB}=alpha`


therefore `angle {APO}=angle {BPO}=frac{alpha}{2}`

In triangle {OAP}, `sin frac{alpha}{2}=frac{{OA}}{{OP}}`


`Rightarrow sin frac{alpha}{2}=frac{r}{O P}`

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