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The radius of the circle passing through the centre of the inscribed circle of the $\Delta$le ABC and through the end point of the base BC is R then $\large\frac{a}{R}$$\sec \large\frac{A}{2} is equal to (a)\;3\qquad(b)\;4\qquad(c)\;2\qquad(d)\;0 1 Answer Comment A) Step 1: Let O be the centre of the inscribed circle of triangle ABC.We have drawn another circle passing through O,B and C.Suppose radius is equal to R. Applying sine rule in \Delta OBC,we get \large\frac{a}{\sin \angle BOC}$$=2R$
$\Rightarrow R=\large\frac{a}{2\sin \angle BOC}$-----(1)
Now,since $O$ is the centre of the inscribed circle,hence $BO$ and $CO$ are bisections of angles $B$ and $C$ respectively.
$\angle OBC=\large\frac{B}{2}$
$\angle OCB=\large\frac{C}{2}$
$\angle BOC=180^{\large\circ}-\big(\large\frac{B}{2}+\frac{C}{2}\big)$
$\qquad\;\;=180^{\large\circ}-\big(90^{\large\circ}-\large\frac{A}{2}\big)$
$\qquad\;\;=90^{\large\circ}+\large\frac{A}{2}$
Step 2:
Substituting the value in (1) we get,
$R=\large\frac{a}{2\sin (90^{\large\circ}+\large\frac{A}{2}\big)}$
$\;\;=\large\frac{a}{2}$$\sec\large\frac{A}{2} \;\;=\large\frac{a}{R}$$\sec\large\frac{A}{2}$
$\;\;=2$
Hence (c) is the correct answer.