Answer
$a\approx6.3, B\approx51^\circ, C\approx27^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$a^2=b^2+c^2-2bc\ cosA$, we have $a^2=(5)^2+(3)^2-2(5)(3)cos(102^\circ)\approx 40.24$, thus $c=\sqrt {40.24}\approx 6.3$
Step 2. Using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{5}{6.3}sin(102^\circ)\approx 0.7763$, thus $B=sin^{-1}(0.7763)\approx51^\circ$
Step 3. We can find the angle as
$C=180^\circ-102^\circ-51^\circ=27^\circ$
Step 4. We solved the triangle with
$a\approx6.3, B\approx51^\circ, C\approx27^\circ$
