Prove that, if 2 angles of a spherical triangle are equal, then the triangle is an isosceles spherical triangle

Question

So the question goes: "An Isosceles Spherical Triangle is a triangle that has 2 sides of equal length. Prove that, if 2 angles of a spherical triangle are equal, then the triangle is an isosceles spherical triangle."

I've only started learning spherical trigonometry a few days ago so I am still very new to all of this. I've learnt the cosine and sine theorems, but not sure how to use them to prove this. I'd love for some help and ideally if you do decide to respond to this, please try to keep it simple as my knowledge is quite limited. Thanks in advance!

Answer 1

Assume $A=B=X$ (look hier for definitions), then we need to show that $a=b$. Using the sine rule, we have $$\sin(A) / \sin(a) =\sin(X) / \sin(a) = \sin(X)/\sin(b)=\sin(B) / \sin(b)$$ or $$Y =\color{red}{\sin(a) = \sin(b)}$$ We also could use the cosine rules $$\cos(a) = \cos(b)\cos(c)+\sin(b)\sin(c)\cos(A)$$ $$\cos(b) = \cos(c)\cos(a)+\sin(a)\sin(c)\cos(B)$$ or $$\cos(a) = \cos(b)\cos(c)+Y\sin(c)\cos(X)$$ $$\cos(b) = \cos(a)\cos(c)+Y\sin(c)\cos(X),$$ because $X=A=B$ and $Y=\sin(a) = \sin(b)$.

Subtracting correctly gives $(\cos(a)-\cos(b))(1+\cos(c))=0$. This leads to $\color{red}{\cos(a) = \cos(b)}$ or $\cos(c)=-1$.

If $\color{red}{\cos(a) = \cos(b)}$, then with $\color{red}{\sin(a) = \sin(b)}$ we are done and $a=b$.

Otherwise if $\cos(a)\neq\cos(b) $ and $\color{red}{\cos(c)=-1}$, then with $\color{red}{\sin(a) = \sin(b)}$...

Answer 2

This is not a problem in trigonometry; rather the proof is the same in spherical geometry as it is in Euclidean:

Let the two angles be at $A$ and $B$, the third vertex being called $C$. Now draw a segment from $C$ to the midpoint $M$ of $\overline{AB}$. Then you have two triangles $\triangle CMA$ and $\triangle CMB$, congruent by SAS.