# On finding the values of angles using sine and cosine rules

I had a triangle $$\triangle ABC$$, given that, $$BC=\sqrt{3}+1$$, $$AC = \sqrt{3}-1$$, and $$\angle BCA=60^\circ$$. I am asked to find the value of $$\angle BAC$$.

So I used to cosine rule to get $$AB = \sqrt{6}$$, and then tried using the sine rule to get to the value of $$\angle BAC$$.

I got $$\sin(A)=\frac{\sqrt3+1}{2\sqrt2}$$ For simplicity, I am writing $$\angle BAC$$ as $$\angle A$$

Using a calculator, that evaluates to be

$$\angle A = \arcsin(0.965925826)$$

According to my knowledge and calculator, that can be both $$75^\circ$$ or $$105^\circ$$. What should be my answer if both options are given in the question?

• There seems to be a typo in the question. Why do you need to find $\angle BAC$ when it is already given? In any case, you can always use the sine rule to find out the third angle. That should give you an idea (sum of the angles should be 180). Commented Oct 28, 2019 at 7:39
• @Iguana Corrected the typo. Thanks Commented Oct 28, 2019 at 7:54

You already got $$\sin\angle{BAC}=\frac{\sqrt 3+1}{2\sqrt 2}$$ which can be written as $$\sin{\angle{BAC}}=\frac{1}{\sqrt 2}\cdot\frac{\sqrt 3}{2}+\frac{1}{\sqrt 2}\cdot \frac 12=\sin(45^\circ+30^\circ)$$ which implies that
$$\angle{BAC}=75^\circ\quad\text{or}\quad 105^\circ$$
Now note that $$BC^2-AB^2-CA^2=2(2\sqrt 3-3)\gt 0$$which implies that $$\angle{BAC}$$ is obtuse, so $$\angle{BAC}=\color{red}{105^\circ}$$.
You can write $$c^2=a^2+b^2-2ab\cos(\gamma)$$ then $$\frac{\sin(\gamma)}{\sin(\pi/3)}=\frac{c}{a}$$ so $$\frac{a^2\sin(\gamma)^2}{\sin(\pi/3)^2}=a^2+b^2-2ab\cos(\gamma)$$ for $$\sin(\gamma)^2$$ we can write $$1-\cos(\gamma)^2$$ and you will get a quadratic equation for $$\gamma$$