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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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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

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

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1 Answer

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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.
answered Nov 19, 2013 by sreemathi.v
edited Mar 5, 2014 by meenakshi.p
 

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