Q. In a $\Delta ABC, \frac{a}{b} = 2+\sqrt{3}$ and $∠C = 60^{\circ}$. Then the ordered pair $\left(∠A, ∠B\right)$ is equal to :

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A

$(15^{\circ}$, $105^{\circ}$)

14%

B

$(105^{\circ}, 15^{\circ}$)

43%

C

$(45^{\circ}, 75^{\circ}$)

29%

D

$(75^{\circ}, 45^{\circ}$)

14%

Solution:

$\frac{a}{b}=2-\sqrt{3}, \angle C=60^{\circ}$.
$\frac{a}{b}=\frac{\sin A}{\sin B}=(2-\sqrt{3})=\tan 15^{\circ}=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}$
Now, since $\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)=\frac{1-1 / \sqrt{3}}{1+1 / \sqrt{3}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}$
$\Rightarrow \frac{\sin A}{\sin B}=\frac{\sin 15^{\circ}}{\sin \left(90^{\circ}+15^{\circ}\right)}$
$=\frac{\sin 15^{\circ}}{\sin 105^{\circ}}$
Also $\angle C=60^{\circ}$
$\Rightarrow \angle A+\angle B=120^{\circ}$
$\Rightarrow \angle A=15^{\circ}, $
$\angle B=105^{\circ}$

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