Answer
$A= 18^\circ17'$, $B= 71^\circ43'$, $b= 14.5\ m$
Work Step by Step
Use the definitions of sine, we have $sin(A)=\frac{4.80}{15.3}$ which gives $A=asin(\frac{4.80}{15.3})\approx18^\circ17'$, thus angle $B=90^\circ-18^\circ17'=71^\circ43'$.
Use Pythagorean's Theorem, we have $b=\sqrt {(15.3)^2-(4.80)^2}\approx14.5\ m$ (or use $b=15.3cosA$)
