In a triangle A B C , a b = 2 + 3 and ∠ C = 60 ° . Then the measure of ∠ A is

Given:

$\frac{a}{b}=2+\sqrt{3}$

$\angle C=60°$

Also,

$A+B+C=180°$

$\Rightarrow A+B=120° ...\left(1\right)$

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

$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{\sqrt{3}+3}{\sqrt{3}+1}$

$\Rightarrow \frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}=\frac{3+\sqrt{3}}{\sqrt{3}+1}$

$\Rightarrow \frac{\sin \left(60°\right)\cos \left(\frac{A-B}{2}\right)}{\sin \left(\frac{A-B}{2}\right)\cos \left(60°\right)}=\frac{\sqrt{3}\left(1+\sqrt{3}\right)}{\sqrt{3}+1}$

$\Rightarrow \sqrt{3}\cot \left(\frac{A-B}{2}\right)=\sqrt{3}$

$\Rightarrow \cot \left(\frac{A-B}{2}\right)=1$

$\Rightarrow \frac{A-B}{2}=45°$

$\Rightarrow A-B=90° ...\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get

$A=105°, B=15°$