With the usual notation,in Delta ABC,if angle | vedclass

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(A) Given $\angle A + \angle B = 120^{\circ}$ and $\frac{a}{b} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.Using the Sine Rule,$\frac{\sin A}{\sin B} = \frac{a}{

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$7 : 1$

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(A) Given $\angle A + \angle B = 120^{\circ}$ and $\frac{a}{b} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.Using the Sine Rule,$\frac{\sin A}{\sin B} = \frac{a}{b} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.Since $\angle B = 120^{\circ} - A$,we have $\frac{\sin A}{\sin(120^{\circ}-A)} = 2+\sqrt{3}$.$\frac{\sin A}{\sin 120^{\circ} \cos A - \cos 120^{\circ} \sin A} = 2+\sqrt{3}$.$\frac{\sin A}{\frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A} = 2+\sqrt{3}$.Dividing numerator and denominator by $\sin A$,we get $\frac{1}{\frac{\sqrt{3}}{2} \cot A + \frac{1}{2}} = 2+\sqrt{3}$.$\frac{2}{\sqrt{3} \cot A + 1} = 2+\sqrt{3} \implies \sqrt{3} \cot A + 1 = \frac{2}{2+\sqrt{3}} = 2(2-\sqrt{3}) = 4-2\sqrt{3}$.$\sqrt{3} \cot A = 3-2\sqrt{3} \implies \cot A = \sqrt{3}-2$.Since $\cot 105^{\circ} = \cot(60^{\circ}+45^{\circ}) = \frac{\cot 60^{\circ} \cot 45^{\circ}-1}{\cot 60^{\circ}+\cot 45^{\circ}} = \frac{\frac{1}{\sqrt{3}}-1}{\frac{1}{\sqrt{3}}+1} = \frac{1-\sqrt{3}}{1+\sqrt{3}} = \frac{(1-\sqrt{3})^2}{1-3} = \frac{4-2\sqrt{3}}{-2} = \sqrt{3}-2$.Thus,$A = 105^{\circ}$ and $B = 120^{\circ} - 105^{\circ} = 15^{\circ}$.The ratio $A : B = 105^{\circ} : 15^{\circ} = 7 : 1$.

With the usual notation,in $\Delta ABC$,if $\angle A + \angle B = 120^{\circ}$ and $a : b = (\sqrt{3} + 1) : (\sqrt{3} - 1)$,then the ratio $\angle A : \angle B$ is

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