In a triangle ABC,if sin frac{A}{2} = frac{1} | vedclass

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(B) Given $\sin^2 \frac{A}{2} = \frac{1}{16} 	imes \frac{3}{5} = \frac{3}{80}$.Using the formula $\cos A = 1 - 2 \sin^2 \frac{A}{2} = 1 - 2(\frac{3}{

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$\frac{\sqrt{231}}{4}$

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(B) Given $\sin^2 \frac{A}{2} = \frac{1}{16} 	imes \frac{3}{5} = \frac{3}{80}$.Using the formula $\cos A = 1 - 2 \sin^2 \frac{A}{2} = 1 - 2(\frac{3}{80}) = 1 - \frac{3}{40} = \frac{37}{40}$.By the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$.$2^2 = b^2 + 5^2 - 2(b)(5)(\frac{37}{40})$.$4 = b^2 + 25 - \frac{37b}{4}$.$16 = 4b^2 + 100 - 37b$.$4b^2 - 37b + 84 = 0$.Solving for $b$ using the quadratic formula: $b = \frac{37 \pm \sqrt{37^2 - 4(4)(84)}}{2(4)} = \frac{37 \pm \sqrt{1369 - 1344}}{8} = \frac{37 \pm \sqrt{25}}{8} = \frac{37 \pm 5}{8}$.$b = \frac{42}{8} = 5.25$ or $b = \frac{32}{8} = 4$.Since $b$ is an integer,$b = 4$.The area of triangle $ABC = \frac{1}{2} bc \sin A$.We have $\cos A = \frac{37}{40}$,so $\sin A = \sqrt{1 - (\frac{37}{40})^2} = \sqrt{\frac{1600 - 1369}{1600}} = \frac{\sqrt{231}}{40}$.Area $= \frac{1}{2} 	imes 4 	imes 5 	imes \frac{\sqrt{231}}{40} = 10 	imes \frac{\sqrt{231}}{40} = \frac{\sqrt{231}}{4}$.

In a triangle $ABC$,if $\sin \frac{A}{2} = \frac{1}{4} \sqrt{\frac{3}{5}}$,$a = 2$,$c = 5$ and $b$ is an integer,then the area (in sq. units) of triangle $ABC$ is

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