Answer
$\text{Area}\approx14.34457\,cm^{2}$
Work Step by Step
Using the law of cosines, we have
$(|CA|)^{2}=(|AB|)^{2}+(|BC|)^{2}-2(|AB|)(|BC|)\cos B$
$=(10\,cm)^{2}+(3\,cm)^{2}-2(10\,cm)(3\,cm)\cos 107^{\circ}$
$\approx126.542302\,cm^{2}$
$|CA|\approx11.2491023\,cm$
The semiperimeter of the triangle is
$s=\frac{1}{2}(10\,cm+3\,cm+11.2491023\,cm)$
$=12.1245512\,cm$
$\text{Area}=\sqrt {s(s-a)(s-b)(s-c)}$ where $a,b $ and $c$ are the lengths of the sides of the triangle.
$\text{Area}=\sqrt {12.1245512(2.1245512)(9.1245512)(0.8754489)}cm^{2}$
$\approx14.34457\,cm^{2}$
