Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Find other angles

Question

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 ?

Answer 1

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.

Answer 2

By law of cosines we obtain: $$\cos\alpha=\frac{(\sqrt3+1)^2+(\sqrt6)^2-(\sqrt3-1)^2}{2(\sqrt3+1)\sqrt6}=\frac{\sqrt3+1}{2\sqrt2},$$ which gives $\alpha=15^{\circ}$ and from here $\beta=105^{\circ}.$

Answer 3

Playing around with the angle sum and the Law of Sines ($|\angle A|>|\angle B|$):

$|\angle A|+|\angle B|=120°$

$(\sin A)/(\sin B)=(\sqrt{3}+1)/(\sqrt{3}-1)=2+\sqrt{3}$

Put $|\angle A|=120°-|\angle B|$ and apply the formula for the wine of a difference:

$((\sqrt{3}/2)(\cos B)-(1/2)(\sin B))/(\sin B)=2+\sqrt{3}$

$(\cos B)/(\sin B)=\cot B = 2+\sqrt{3}$

$\tan B=2-\sqrt{3}$

Then

$\tan A = \tan (120°-B)=\dfrac{-\sqrt{3}-(2-\sqrt{3})}{1+(-\sqrt{3})((2-\sqrt{3}))}=-(2+\sqrt{3})$

Then $\tan A \tan B=-1$ so the angle measures must differ by a right angle. So

$|\angle A|+|\angle B|=120°$

$|\angle A|-|\angle B|=90°$

$|\angle A|=105°, |\angle B|=15°$