By the law of sines, we have
[tex]\dfrac{\sin\theta}{12.3}=\dfrac{\sin B'}{13}=\dfrac{\sin B}{13}[/tex]
You might be tempted to conclude that [tex]B=B'[/tex], but the picture clearly indicates otherwise.
The triangle on the left containing angle [tex]B[/tex] is isosceles, which tells us that the angle directly adjacent to [tex]B'[/tex] is congruent to angle [tex]B[/tex] and thus has the same measure. Angles [tex]B[/tex] and [tex]B'[/tex] are therefore supplementary and
[tex]m\angle B=180^\circ-m\angle B'[/tex]
It's also true that
[tex]\sin x=\sin(180^\circ-x)[/tex]
for any angle [tex]x[/tex], so there's no issue here and it's perfectly fine that the equation above has two possible solutions.
So all we have to do is solve for one of those angles and we get the other for free by adding/subtracting 180 accordingly.
[tex]\dfrac{\sin\theta}{12.3}=\dfrac{\sin B}{13}\implies\sin B=\dfrac{13}{12.3}\sin60^\circ\approx0.9153[/tex]
[tex]\implies B\approx\sin^{-1}0.9153=66.25^\circ[/tex]
Then [tex]B'=180^\circ-B=113.75^\circ[/tex].
