Finding the length of the sides, the measure of the angles and area of spherical triangles?

Question

I'm trying to understand this problem in the textbook but I got lost in one part:

Problem: Assume that the earth is a sphere of radius $5280$ miles, find the length of the sides, the measure of the angles and the area of the spherical triangle with vertices $A(70° N, 10° E), B(10° S, 100° E)$ and $C(50° S,80° W)$. The earth's radius will be used as the unit of length. The spherical coordinates $(r,v,u)$ of the three vertices are $(1,10,20),(1,100,100) $ and $(1,-80,140)$.

Their cartesian coordinates are:

\begin{align*} (\sin(20°)\cos(10°), \sin(20°)\sin(10°),\cos(20°))&=(0.3368,0.0594,0.9397)\\ (\sin(100°)\cos(100°), \sin(100°)\sin(100°),\cos(100°))&=(-0.1710,0.9698,-.1736)\\ (\sin(140°)\cos(-80°), \sin(140°)\sin(-80°),\cos(140°))&=(0.1116,-0.6330,-.7660) \end{align*}

THIS IS WHERE I BEGIN TO HAVE PROBLEMS UNDERSTANDING:

The cosines of the angles between radii $OA$, $OB$ and $OC$ are equal to the dot product of the corresponding position vectors. Thus

\begin{align*} \angle AOB&=\cos^{-1}(-0.1631)=1.7347\text{ radians}\\ \angle BOC&=\cos^{-1}(-0.5000)=2.0944\text{ radians}\\ \angle COA&=\cos^{-1}(-0.7198)=2.3743\text{ radians} \end{align*}

Where did they get those numbers? As in numbers, I mean $-0.1631,-0.5000,-0.7198$

Answer 1

These numbers are simply the dot products.

\begin{align*}\cos\angle AOB&=\frac{\overrightarrow{OA}\cdot\overrightarrow{OB}}{\lvert OA\rvert\cdot\lvert OB\rvert}\\&=\frac{0.3368\cdot(-0.1710)+0.0594\cdot0.9698+0.9397\cdot(-0.1736)}{1\cdot 1}\\&=-0.1631186\approx-0.1631\end{align*}