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In a $\Delta ABC, \frac{a}{b}=2+\sqrt{3}\: \: and\angle C=60^{\circ}$

Then the ordered pair $\left ( \angle A,\angle B \right )$ is equal to
Option: 1
$\left ( 15^{\circ} ,105^{\circ}\right )$

Option: 2
$\left ( 105^{\circ} ,15^{\circ}\right )$

Option: 3
$\left ( 45^{\circ} ,75^{\circ}\right )$

Option: 4
$\left ( 75^{\circ} ,45^{\circ}\right )$

Answers (1)

best_answer

$\frac{\sin A}{\sin B}= \frac{a}{b} \: \: \: \: \: \left [ \sin law \right ]$

$\Rightarrow \frac{\sin A}{\sin B}= 2+\sqrt{3} \: \: \: \cdot \cdot \cdot (1)$

Also, $A+B+C=180^{\circ}\Rightarrow A+B=180^{\circ}-C\:\\ \Rightarrow A+B=120^{\circ} \: \: \: \cdot \cdot \cdot (2)$

From (1) and (2), $\frac{\sin \left ( 120-B \right )}{\sin B}=2+\sqrt{3}$

$\Rightarrow \frac{\sin 120\cos B-\sin B\cos 120}{\sin B}=2+\sqrt{3}$

$\Rightarrow \frac{\sqrt{3}}{2}\cot B+\frac{1}{2}=2+\sqrt{3}$

$\Rightarrow \frac{\sqrt{3}}{2}\cot B=\frac{3}{2}+\sqrt{3}$

$\Rightarrow \cot B=2+\sqrt{3}$

$\therefore B=15^{\circ}$

$A=105^{\circ}$

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