Answer
$\angle A = 70^{\circ} 10'$
$\angle B = 19^{\circ} 50'$
$a\approx 609m$
Work Step by Step
1. We know $b=219m$ and $c = 647m$
2. We use inverse sine to find $\angle B$
$\angle B = \sin ^{-1} \frac{219}{647} \approx 19.8^{\circ} \approx 19^{\circ}50'$ minutes are rounded for three significant figures
3. We use cosine to find side $a$
$\cos 19.8^{\circ} = \frac{a}{647}$
$a\approx 609m$
4. Angles in a triangle must add up to $180^{\circ}$, therefore
$\angle A = 180^{\circ} -90^{\circ} -19.8^{\circ} = 70.2 ^{\circ} = 70^{\circ} 10'$
