Two sides of a triangle are sqrt{3}+1 and sqr | vedclass

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(D) Let $a = \sqrt{3}+1$,$b = \sqrt{3}-1$,and $C = 60^{\circ}$.Using the Law of Cosines:$c^2 = a^2 + b^2 - 2ab \cos C$$c^2 = (\sqrt{3}+1)^2 + (\sqrt{3

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(D) Let $a = \sqrt{3}+1$,$b = \sqrt{3}-1$,and $C = 60^{\circ}$.Using the Law of Cosines:$c^2 = a^2 + b^2 - 2ab \cos C$$c^2 = (\sqrt{3}+1)^2 + (\sqrt{3}-1)^2 - 2(\sqrt{3}+1)(\sqrt{3}-1) \cos 60^{\circ}$$c^2 = (3+1+2\sqrt{3}) + (3+1-2\sqrt{3}) - 2(3-1) 	imes \frac{1}{2}$$c^2 = 8 - 2 = 6 \implies c = \sqrt{6}$.Using the Law of Tangents:$	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$$	an\left(\frac{A-B}{2}ight) = \frac{(\sqrt{3}+1) - (\sqrt{3}-1)}{(\sqrt{3}+1) + (\sqrt{3}-1)} \cot(30^{\circ})$$	an\left(\frac{A-B}{2}ight) = \frac{2}{2\sqrt{3}} 	imes \sqrt{3} = 1$.Therefore,$\frac{A-B}{2} = 45^{\circ} \implies A-B = 90^{\circ}$.

Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Find the difference of the remaining angles. (in $^{\circ}$)

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