Spherical Pythagorean Theorem In class, we came across the following relation for a right triangle on the surface of the sphere. $cos(\frac{c}{R})=cos(\frac{a}{R})cos(\frac{b}{R})$ where R is the radius of the sphere. Here a, b, c are sides of the triangle opposite to their respective angles $\alpha$, $\beta$, and $\gamma=\frac{\pi}{2}$. I don't quite understand why this works. My understanding is that on the surface of the sphere $c=\gamma R$, $b=\beta R$, and $a=\alpha R$. Thus the angles which subtend the arc are equal to the fractions in the formula. However if $\gamma=\frac{\pi}{2}$rad then we would get that $0=cos(\frac{a}{R})cos(\frac{b}{R})$. Would this not force either of $\alpha$ or $\beta$ to be $\frac{\pi}{2}$rad? Does this mean that on a sphere, if one angle is $\frac{\pi}{2}$rad then at least one more is as well? If I am missing anything please let me know. Also, please try to make any answers as simple as possible because I am new at this.
Thanks for any feedback
 A: It’s really quite easy, once you draw the right picture.
After that, you have to see that the lengths of the triangle’s sides are best measured as angles, namely the angle subtended by each as seen from the sphere’s center.
Now put your triangle on the surface of the unit sphere that’s centered at the origin of $(x,y,z)$-space, with the right angle of your triangle on the $x$-axis, at $(1,0,0)$, one side sitting in the $x,z$-plane, say that’s your side $A$, the other side $B$ sitting in the horizontal plane, that’s the $x,y$-plane.
Do you have it all down on paper now? You check that the other end of the $A$ side is at the point $(\cos A,0,\sin A)$, and the other end of the $B$-side is at $(\cos B,\sin B,0)$.
Now you have two vectors in space, and you realize that the hypotenuse of your triangle runs from one vector’s endpoint to that of the other. What angle measures it? The cosine of that angle is the dot product of the two vectors, divided by the lengths, which are fortunately $1$.  But the dot product of these two vectors is $\cos A\cos B$. And there’s your spherical Pythagorean Theorem.
