In triangle ABC,if a=2,B=tan ^{-1} frac{1}{2} | vedclass

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(C) Given that,$a=2$. In $	riangle ABC$,$B=	an ^{-1}(\frac{1}{2})$ and $C=	an ^{-1}(\frac{1}{3})$. We know that in $	riangle ABC$,$A+B+C=\pi$. $A

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$\frac{3 \pi}{4}, \frac{2 \sqrt{2}}{\sqrt{5}}$

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(C) Given that,$a=2$. In $	riangle ABC$,$B=	an ^{-1}(\frac{1}{2})$ and $C=	an ^{-1}(\frac{1}{3})$. We know that in $	riangle ABC$,$A+B+C=\pi$. $A = \pi - (B+C) = \pi - (	an ^{-1}(\frac{1}{2}) + 	an ^{-1}(\frac{1}{3}))$. Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1}(\frac{x+y}{1-xy})$,we get: $B+C = 	an ^{-1}(\frac{1/2 + 1/3}{1 - 1/6}) = 	an ^{-1}(\frac{5/6}{5/6}) = 	an ^{-1}(1) = \frac{\pi}{4}$. Thus,$A = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Now,$\sin A = \sin(\frac{3\pi}{4}) = \sin(135^{\circ}) = \frac{1}{\sqrt{2}}$. Since $	an B = \frac{1}{2}$,we have $\sin B = \frac{1}{\sqrt{1^2+2^2}} = \frac{1}{\sqrt{5}}$. Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B}$. $b = a \cdot \frac{\sin B}{\sin A} = 2 \cdot \frac{1/\sqrt{5}}{1/\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{5}}$. Therefore,$(A, b) = (\frac{3\pi}{4}, \frac{2\sqrt{2}}{\sqrt{5}})$.

In $\triangle ABC$,if $a=2$,$B=\tan ^{-1} \frac{1}{2}$ and $C=\tan ^{-1} \frac{1}{3}$,then $(A, b)$ equals

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