In a triangle ABC,frac{a}{b}=2+sqrt{3} and an | vedclass

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(C) Given: $\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$.Since $A+B+C=180^{\circ}$,we have $A+B=120^{\circ}$ $(1)$.By the sine rule,$\frac{\sin A

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$105$

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(C) Given: $\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$.Since $A+B+C=180^{\circ}$,we have $A+B=120^{\circ}$ $(1)$.By the sine rule,$\frac{\sin A}{\sin B} = \frac{a}{b} = 2+\sqrt{3}$.Applying componendo and dividendo:$\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2+\sqrt{3}+1}{2+\sqrt{3}-1} = \frac{3+\sqrt{3}}{\sqrt{3}+1} = \frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = \sqrt{3}$.Using the sum-to-product formulas:$\frac{2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})}{2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})} = \sqrt{3}$.$	an(\frac{A+B}{2}) \cot(\frac{A-B}{2}) = \sqrt{3}$.Since $\frac{A+B}{2} = 60^{\circ}$,$	an(60^{\circ}) \cot(\frac{A-B}{2}) = \sqrt{3}$.$\sqrt{3} \cot(\frac{A-B}{2}) = \sqrt{3} \Rightarrow \cot(\frac{A-B}{2}) = 1$.Thus,$\frac{A-B}{2} = 45^{\circ} \Rightarrow A-B = 90^{\circ}$ $(2)$.Adding $(1)$ and $(2)$: $2A = 210^{\circ} \Rightarrow A = 105^{\circ}$.

In a triangle $ABC$,$\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$. Then the measure of $\angle A$ is (in $^{\circ}$)

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