Question

In a $\Delta ABC$, if $\frac{a}{b}=2+\sqrt{3}$ and $\angle C={60}^o$ show that $\angle A={105}^o$ and $\angle B={15}^o$

Answer 1

$\frac{a}{b}=2t\sqrt{3}\angle c={60}^o$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin c}{C}$

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

$A+B+C=180$

$A+B=120\Rightarrow B=120-A$

$\frac{\sin A}{\sin (120-A)}=2+\sqrt{3}$

$\tan 75$

$\frac{\sin A}{\sin (120-A)}=\frac{\sin {75}^o}{\cos {75}^o}=\frac{\sin {75}^o}{\sin {15}^o}$

$A={105}^o,\angle c={60}^o$

$B={15}^o$