Question: A Spherical triangle has angles of $120^{\circ}$, $60^{\circ}$ and $45^{\circ}$. Find the cosines of the (arc) lengths of the sides. How many sides have an arc length greater than $90^{\circ}$?
I applied the law of cosine which is $$\cos (a)=\frac{\cos (\alpha) +\cos(\beta) \cos (\gamma)}{\sin (\beta) \sin (\gamma)}$$
I actually got the answer but I wanted to make clear some concepts. The law of cosine with the plus sign, which is the one above is only applied when one of the angles is greater than the quadrant. So the quadrant has to be greater than $90^{\circ}$ and since there is an angle of $120^{\circ}$ then we use :
$$\cos (a)=\frac{\cos (\alpha) +\cos(\beta) \cos (\gamma)}{\sin (\beta) \sin (\gamma)}$$ instead of :
$$\cos (a)=\frac{\cos (\alpha) -\cos(\beta) \cos (\gamma)}{\sin (\beta) \sin (\gamma)}$$
Is that right?
No, the law of cosine with a plus sign is for when you have more angles or all angles and you are looking for side. The one with the minus sign is when you have more sides or all sides and looking for angles.