Answer
$A= 42^\circ07'$, $a= 270.0\ m$, and $c= 402.5\ m$
Work Step by Step
Given $B=47^\circ53', b=298.6\ m$, we have $A=90^\circ-47^\circ53'=42^\circ07'$, $a=\frac{b}{tanB}=\frac{298.6}{tan47^\circ53'}\approx270.0\ m$, and $c=\frac{b}{sinB}=\frac{298.6}{sin47^\circ53'}\approx402.5\ m$
