Answer
$\angle A \approx 18.28^{\circ}$
$\angle B \approx 71. 72^{\circ}$
$ b \approx 14.5 m$
Work Step by Step
1. $a = 4.8 m$; $c =15.3 m $
$\angle C = 90^{\circ}$
2. $\sin A = \frac{a}{c}= \frac{4.80}{15.3}$
$\angle A = \sin^{-1} (0.0.3137254902) \approx 18.28^{\circ}$
3. $\angle A+ \angle B = 90^{\circ}$
$ \angle B = 90^{\circ} -18.28^{\circ} = 71.72^{\circ}$
4. Pythagoream theorem
$15.3^2 = 4.80^2 + b^2$
$b = \sqrt {234.09 -23.04}$
$b \approx 14.5 m$ consider only positive root, as length always positive in trigonometry
