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?
Best Answer
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 $\angle{BAC}=\color{red}{105^\circ}$.