With usual notations,in triangle ABC,a=sqrt{3 | vedclass

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(D) Given $a=\sqrt{3}+1$,$b=\sqrt{3}-1$,$m \angle C=60^{\circ}$.Using the tangent rule (Napier's Analogy): $	an \left( \frac{A-B}{2} ight) = \frac{

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(D) Given $a=\sqrt{3}+1$,$b=\sqrt{3}-1$,$m \angle C=60^{\circ}$.Using the tangent rule (Napier's Analogy): $	an \left( \frac{A-B}{2} ight) = \frac{a-b}{a+b} \cot \left( \frac{C}{2} ight)$.Substitute the values: $\frac{a-b}{a+b} = \frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.$	an \left( \frac{A-B}{2} ight) = \frac{1}{\sqrt{3}} \cot \left( \frac{60^{\circ}}{2} ight) = \frac{1}{\sqrt{3}} \cot(30^{\circ}) = \frac{1}{\sqrt{3}} 	imes \sqrt{3} = 1$.$\frac{A-B}{2} = 45^{\circ} \Rightarrow A-B = 90^{\circ}$.

With usual notations,in triangle $ABC$,$a=\sqrt{3}+1$,$b=\sqrt{3}-1$ and $m \angle C=60^{\circ}$,then $A-B=$ (in $^{\circ}$)

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