Answer
Case 1. $B\approx58.5^\circ$, $A\approx 83.0^\circ$, $a \approx1250\ in$
Case 2. $B\approx121.5^\circ$, $A\approx 20.0^\circ$, $a \approx431\ in$
Work Step by Step
1. Given $b=1075\ in, c=785\ in, C=38^\circ30'=38.5^\circ$, use the Law of Sines, we have:
$\frac{sinB}{1075}=\frac{sin38.5^\circ}{785}$ which gives $sinB\approx0.8525$ and $B=asin(0.8525)\approx58.5^\circ$ or $121.5^\circ$
2. For $B\approx58.5^\circ$, we have $A\approx180^\circ-38.5^\circ-58.5^\circ=83^\circ$. Use the Law of Sines again, we have $a=\frac{c\ sinA}{sinC }=\frac{785\ sin83^\circ}{sin38.5^\circ }\approx1250\ in$
3. For $B\approx121.5^\circ$, we have $A\approx180^\circ-38.5^\circ-121.5^\circ=20^\circ$. Use the Law of Sines again, we have $a=\frac{c\ sinA}{sinC }=\frac{785\ sin20^\circ}{sin38.5^\circ }\approx431\ in$
