Answer
$c\approx7.1, A\approx159^\circ, B\approx6^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$c^2=a^2+b^2-2ab\ cosC$, we have $c^2=(10)^2+(3)^2-2(10)(3)cos(15^\circ)\approx51.04$, thus $c=\sqrt {51.04}\approx7.1$
Step 2. Using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinC}{c}$, $SinB=\frac{3}{7.1}sin(15^\circ)\approx0.1094$, thus $B=sin^{-1}(0.1094)\approx6^\circ$
Step 3. We can find the angle as
$A=180^\circ-15^\circ-6^\circ=159^\circ$
Step 4. We solved the triangle with
$c\approx7.1, A\approx159^\circ, B\approx6^\circ$
