A spherical balloon of radius r subtends an a | vedclass

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(A) Let $P$ be the eye of the observer and $B$ be the centre of the spherical balloon. Let $Q$ be a point of tangency on the balloon from $P$. Then $\

Vedclass · Accepted Answer

$r \csc\left(\frac{\alpha}{2}\right) \sin \beta$

Vedclass · Answer

(A) Let $P$ be the eye of the observer and $B$ be the centre of the spherical balloon. Let $Q$ be a point of tangency on the balloon from $P$. Then $\angle QPB = \frac{\alpha}{2}$.In the right-angled triangle $	riangle PQB$,we have $\sin\left(\frac{\alpha}{2}ight) = \frac{BQ}{PB} = \frac{r}{l}$,where $l = PB$ is the distance from the observer to the centre of the balloon.Thus,$l = r \csc\left(\frac{\alpha}{2}ight)$.In the right-angled triangle formed by the height $H$ of the centre $B$ from the ground,we have $\sin \beta = \frac{H}{l}$.Therefore,$H = l \sin \beta = r \csc\left(\frac{\alpha}{2}ight) \sin \beta$.

$A$ spherical balloon of radius $r$ subtends an angle $\alpha$ at the eye of an observer. If the angle of elevation of the centre of the balloon is $\beta$,then the height of the centre of the balloon is:

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