Question

Find the the value of $\angle B$ (in degrees) of the $\Delta ABC$ in which $a=\sqrt{3}+1,b=\sqrt{3}-1,C={60}^{\circ }$

Answer 1

Using Cosine rule in triangle $ABC$,

$c^2=a^2+b^2-2abcosC\Rightarrow c^2=(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-2(\sqrt{3}+1)(\sqrt{3}-1)cos{60}^o=4+4-2=6\Rightarrow c^2=6\Rightarrow c=\sqrt{6}$

and from sine Rule,

$\frac{c}{sinC}=\frac{b}{sinB}\Rightarrow sinB=\frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{6}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\Rightarrow B={15}^o$