Answer
$A= 17.0^\circ$, $a =39.1\ in$, and $c= 134\ in$
Work Step by Step
Given $B=73.0^\circ, b=128\ in$, we have $A=90.0^\circ-73.0^\circ=17.0^\circ$, $a=b\ tanA=128tan(17.0^\circ)\approx39.1\ in$, and $c=\frac{b}{sinB}=\frac{128}{sin(73.0^\circ)}\approx134\ in$
