In triangle ABC,if a=2(sqrt{3}+1),B=45^{circ} | vedclass

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(C) Given $A = 180^{\circ} - (B + C) = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B}

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$6+2 \sqrt{3}$

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(C) Given $A = 180^{\circ} - (B + C) = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.$a = 2(\sqrt{3}+1)$,$\sin 75^{\circ} = \sin(45^{\circ}+30^{\circ}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.So,$\frac{2(\sqrt{3}+1)}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = 4\sqrt{2}$.Thus,$b = 4\sqrt{2} \sin 45^{\circ} = 4\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 4$.$c = 4\sqrt{2} \sin 60^{\circ} = 4\sqrt{2} 	imes \frac{\sqrt{3}}{2} = 2\sqrt{6}$.Area $= \frac{1}{2} ab \sin C = \frac{1}{2} 	imes 2(\sqrt{3}+1) 	imes 4 	imes \sin 60^{\circ} = 4(\sqrt{3}+1) 	imes \frac{\sqrt{3}}{2} = 2(3+\sqrt{3}) = 6+2\sqrt{3}$.

In $\triangle ABC$,if $a=2(\sqrt{3}+1)$,$B=45^{\circ}$ and $C=60^{\circ}$,then the area (in sq.units) of that triangle is

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