• # question_answer With the usual notation, in $\Delta ABC,$ if $\angle A+\angle B=120{}^\circ ,$ $a=\sqrt{3}-1,$ then the ratio $\angle A:\angle B,$ is: A) $7:1$    B) $5:3$ C) $9:7$    D) $3:1$

 $a=\sqrt{3}+1$ $b=\sqrt{3}-1$ $\frac{\sin A}{\sin B}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ $=\frac{3+1+2\sqrt{3}}{2}=2+\sqrt{3}$ $\frac{\sin A}{\sin (120-A)}=\sqrt{3}+2$ $\frac{\sin A}{\sin 12\operatorname{cosA}-cos12sinA}=\sqrt{3}+2$ $\frac{1}{\frac{\sqrt{3}}{2}\cot A+\frac{1}{2}}=\sqrt{3}+2.$
 $\frac{\sqrt{3}\cot A+1}{2}=\frac{1}{\sqrt{3}+2}=\frac{\sqrt{3}-1}{-1}$ $\frac{\sqrt{3}\cot A+1}{2}=-\sqrt{3}+2$ $\sqrt{3}\cot A=4-2\sqrt{3}-1$ $\sqrt{3}\cot A=3-2\sqrt{3}$ $\cot A=\sqrt{3}-2$ $-\cot A=2-\sqrt{3}=\tan 15$ $\therefore$      $A=105{}^\circ$ $\therefore$      $B=15{}^\circ .$