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

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 ?

• Using the Law of Sines to determine an angle can be tricky, since that Law can't distinguish between an angle (for instance, $105^\circ$) and its supplement ($75^\circ$). Try using the Law of Cosines to find $A$. – Blue Dec 30 '18 at 10:36
• BTW: You can use the Law of Sines to find angles opposite the non-longest sides of a triangle. This is because any non-acute angle would be the largest in the triangle and thus also must be opposite the longest side; which is to say, the non-longest sides must be opposite acute angles. Since $\sqrt{3}+1=2.732\ldots$ is larger than both $\sqrt{3}-1=0.732\ldots$ and $\sqrt{6}=2.449\ldots$, its opposite angle happens to be the worst to find with the Law of Sines. – Blue Dec 30 '18 at 11:28

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.

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 $$\beta=105^{\circ}.$$

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

$$|\angle A|+|\angle B|=120°$$

$$|\angle A|-|\angle B|=90°$$

$$|\angle A|=105°, |\angle B|=15°$$