#### Answer

$\angle A = 24^{\circ} 10'$
$\angle B = 65^{\circ} 50'$
$b\approx 42.3 cm$

#### Work Step by Step

1. We know $a=18.9cm$ and $c = 46.3 cm$
2. We use inverse sine to find $\angle A$
$\sin ^{-1} \frac{18.9}{46.3} \approx 24.1^{\circ} \approx 24^{\circ}10'$ minutes are rounded for three significant figures
3. We use cosine to find side $b$
$\cos 24.1^{\circ} = \frac{b}{46.3}$
$b\approx 42.3 cm$
4. Angles in a triangle must add up to $180^{\circ}$, therefore
$\angle B = 180^{\circ} -90^{\circ} -24.1^{\circ} = 65.9 ^{\circ} = 65^{\circ} 50'$