Question

In $\Delta ABC,\frac{a}{b}=2+\sqrt{3}$ and $\angle C={60}^0.$ Then the ordered pair $(\angle A,\angle B)$ is equal to

• A. $({15}^0,{105}^0)$
• B. $({105}^0,{15}^0)$
• C. $({45}^0,{75}^0)$
• D. $({75}^0,{45}^0)$

Answer 1

The correct option is B $({105}^0,{15}^0)$

In $\Delta ABC$
$\angle A+\angle B+\angle C={180}^0$
$\angle A+\angle B+{60}^0={180}^0$
$\angle A+\angle B={120}^0\cdots \left(i\right)$
By sine formula:
$\frac{a}{\sin A}=\frac{b}{\sin B}$
$\frac{\sin A}{\sin B}=\frac{a}{b}$
$\frac{\sin A}{\sin B}=2+\sqrt{3}$
Applying componendo and dividendo, we get
$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{3+\sqrt{3}}{\sqrt{3}+1}$
$\frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}=\frac{3+\sqrt{3}}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$\sqrt{3}\cot \left(\frac{A-B}{2}\right)=\sqrt{3}$
$A-B={90}^0\cdots \left(ii\right)$
Solving equation $\left(i\right),\left(ii\right)$
$A={105}^0,B={15}^0$