Question

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

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

Answer 1

Answer: (A)

(${105}^o,{15}^o$)

We have $A+B+C={180}^0$
$\Rightarrow A+B={120}^0$ ....(1)
We have by sine formula
$\frac{\sin A}{a}=\frac{\sin B}{b}$
$\Rightarrow \frac{\sin A}{\sin B}=\frac{a}{b}$
$\Rightarrow \frac{\sin A}{\sin B}=2+\sqrt{3}$
Applying componendo and dividendo, we get
$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{3+\sqrt{3}}{\sqrt{3}+1}$
$\frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}=\frac{3+\sqrt{3}}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$\Rightarrow \sqrt{3}\cot \left(\frac{A-B}{2}\right)=\sqrt{3}$
$\Rightarrow A-B={90}^0$ .....(2)
Solving (1) and (2), we get
$\Rightarrow A={105}^0,B={15}^0$