How would I go about solving a spherical (right) triangle in the ambiguous case? There are two answers for $m\angle ABC$. I am given the following for:
$m\angle ACB = 90^\circ$, $AC = 2$, $BC = 1$ and $C$ is the midpoint of $AD$.
Using spherical trigonometry of course. I know every right spherical triangle can be solved using Napier's analogies, but what I fail to see is how to get the two angle measures and hence where the ambiguity arises.
From book context, $1\text{ unit}$ means one times the radius. Without loss of generality, assume $r=1$.
We focus on the right spherical triangle $\triangle ABC$ where $C=\angle ACB$ is right, $a=2$ and $b=1$.
Length of hypotenuse $c=\,\stackrel\frown{AB}$ satisfies: $$\cos c= \cos a\cos b =\cos1\cos2 \approx -0.2248.$$
With proper bounding, i.e. $0<c<\pi$, we conclude that $c\approx 1.798$ and $\sin c\approx0.9744$.
Angle $B=\angle ABC$, which is clearly obtuse, satisfies: $$\cos B= \frac{\cos b-\cos a\cos c}{\sin a\sin c}=\frac{\cos 2-\cos 1\times(-0.2248)}{\sin 2\times 0.9744} \approx -0.3326$$
Now, technically, an interior angle of a spherical triangle is only bounded by $0<B<2\pi$, and there are exactly two angles in $[0,2\pi)$ that have a cosine of $-0.3326$. Because our triangle is not that large, we can safely assume that $B\in[0,\pi)$, giving the unique solution of $B\approx 109.42^\circ.$
However, if you tried to obtain $\sin B$ instead of $\cos B$, then you would face a tougher question: Is $B$ acute or obtuse? This is what the author is trying to warn you about. If a careless practitioner went straight for the acute angle, they'll get the wrong answer, because $B$ is obtuse.
Therefore, the meaning of
is to raise awareness of picking the actual quadrant.