IS ASA applicable on triangles on the sphere?

Question

$ASA= \text{Angle-Side-Angle}$

I was wondering if $ASA$ still worked on triangles for the sphere. I have a pretty hard time visualizing triangles on the sphere because I know the sum of their interior angles can be more than $180^{\circ}$, which feels weird to me, since I am accustomed to triangles on the plane. I know $ASA$ works on triangles on the plane, as you can just find the third angle by subtracting the sum of the other angles from $180^{\circ}$, and then use the Law of Sines to find out the other side lengths. Would $ASA$ work on a triangle on a sphere? Thanks in advance.

Answer 1

Indeed. Knowing a side ($\gamma$) and the two angles at either side ($A$ and $B$), we can use the dual Law of Cosines to get the third angle ($C$): $$ \cos(C)=-\cos(A)\cos(B)+\sin(A)\sin(B)\cos(\gamma) $$ Then, we can use the Law of Sines as usual to get the other sides.

Alternatively, we can use the same dual Law of Cosines to get $$ \cos(\alpha)=\frac{\cos(A)+\cos(B)\cos(C)}{\sin(B)\sin(C)} $$ and $$ \cos(\beta)=\frac{\cos(B)+\cos(A)\cos(C)}{\sin(A)\sin(C)} $$ Note that, unlike plane trigonometry, we can determine a spherical triangle by AAA.

Answer 2

If you allow degenerate triangles, then no, otherwise yes.

If you try to draw edges at appropriate angles from the known side, then their intersection is almost always uniquely determined, and so the whole triangle is. Angles and side lenghts are invariant under rotations, so you can actually "move" one triangle to the other so that their known side matches, and then the other triangle edges have to follow the same great circles, thus having the same intersection point.

However, consider a case where your side is of length $\pi$ and the angles are $\frac{\pi}{2}$ – you know how the triangle looks like, but you don't know where the third point actually is. It is similar to when both angles are $0$, but here the shape can have actually non-zero area (which is different from the planar case).

This feels much like nit-picking, but nevertheless I hope it helps $\ddot\smile$