Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Then find the other angles
My Attempt
Let $a=\sqrt{3}+1$, $b=\sqrt{3}-1$ and $C=60$
$$
c^2=a^2+b^2-2.a.b\cos C\\=(\sqrt{3}+1)^2
+(\sqrt{3}-1)^2-2(\sqrt{3}+1)(\sqrt{3}-1).\frac{1}{2}
=8-2=6\\
\implies c=\sqrt{6}=\sqrt{2}\sqrt{3}\\
\frac{a}{\sin A}=\frac{c}{\sin C}\implies\sin A=\frac{\sqrt{3}+1}{\sqrt{2}\sqrt{3}}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}\\
A=75^\circ\quad\&\quad B=45^\circ
$$
But the solution given in my reference is $105^\circ$ and $15^\circ$, what is going wrong with my attempt ?
Best Answer
Because $\sin(\alpha) = \sin(180-\alpha)$. Hence, $\sin(75^\circ) = sin(105^\circ)$. Therefore you should compute the value of $\sin B$ instead to specify one of them. Probably, when you computing the value of angle $B$ you will get $165^\circ$ and $15^\circ$, but as the sum of angles is equal to $180^\circ$, $15^\circ$ will be accepted.