$(a)\;3\qquad(b)\;4\qquad(c)\;2\qquad(d)\;0$

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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.

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