Let PQ and RS be tangents at the extremeties | vedclass

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Question

(a) $	an 	heta = \frac{{PQ}}{{PR}} = \frac{{PQ}}{{2r}}$

Also $	an \left( {\frac{\pi }{2} - 	heta } ight) = \frac{{RS}}{{2r}}$

$i.e.$, $\co

Vedclass · Accepted Answer

$\sqrt {PQ.RS} $

Vedclass · Answer

(a) $	an 	heta = \frac{{PQ}}{{PR}} = \frac{{PQ}}{{2r}}$

Also $	an \left( {\frac{\pi }{2} - 	heta } ight) = \frac{{RS}}{{2r}}$

$i.e.$, $\cot 	heta = \frac{{RS}}{{2r}}$

$	an 	heta .\cot 	heta = \frac{{PQ.RS}}{{4{r^2}}}$

==> $4{r^2} = PQ.RS$

==> $2r = \sqrt {(PQ)(RS)} $.

Let $PQ$ and $RS$ be tangents at the extremeties of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $2r$ equals

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