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

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(D) The area of $	riangle ABC$ is given by $\Delta = \frac{1}{2} ac \sin B$. Given $\Delta = 6+2\sqrt{3}$,$a = 2(\sqrt{3}+1)$,and $B = 45^{\circ}$. S

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$4$

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(D) The area of $	riangle ABC$ is given by $\Delta = \frac{1}{2} ac \sin B$. Given $\Delta = 6+2\sqrt{3}$,$a = 2(\sqrt{3}+1)$,and $B = 45^{\circ}$. Substituting these values: $6+2\sqrt{3} = \frac{1}{2} 	imes 2(\sqrt{3}+1) 	imes c 	imes \sin 45^{\circ}$. $6+2\sqrt{3} = (\sqrt{3}+1) 	imes c 	imes \frac{1}{\sqrt{2}}$. $c = \frac{\sqrt{2}(6+2\sqrt{3})}{\sqrt{3}+1} = \frac{2\sqrt{2}(3+\sqrt{3})}{\sqrt{3}+1} = \frac{2\sqrt{2}\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = 2\sqrt{6}$. Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$. $b^2 = [2(\sqrt{3}+1)]^2 + (2\sqrt{6})^2 - 2[2(\sqrt{3}+1)](2\sqrt{6}) \cos 45^{\circ}$. $b^2 = 4(3+1+2\sqrt{3}) + 24 - 8\sqrt{6}(\sqrt{3}+1) 	imes \frac{1}{\sqrt{2}}$. $b^2 = 16 + 8\sqrt{3} + 24 - 8\sqrt{3}(\sqrt{3}+1) = 40 + 8\sqrt{3} - 24 - 8\sqrt{3} = 16$. Therefore,$b = \sqrt{16} = 4$.

If in $\triangle ABC$,$B=45^{\circ}$,$a=2(\sqrt{3}+1)$ and the area of $\triangle ABC$ is $6+2\sqrt{3}$ sq. units,then the side $b=$

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