#### Answer

$46.4$ m$^{2}$

#### Work Step by Step

The area of the triangle is half the product of the length of two sides and the sine of the angle included between them:
$Area=\frac{1}{2}bc \sin A$
We substitute the values of $A,b$ and $c$ in this formula and solve:
$Area=\frac{1}{2}bc \sin A$
$Area=\frac{1}{2}(13.6)(10.1) \sin 42.5^{\circ}$
$Area=\frac{1}{2}(137.36) \sin 42.5^{\circ}$
$Area=68.68 \sin 42.5^{\circ}$
Using a calculator, $\sin 42.5^{\circ}=0.6756$. Therefore,
$Area=68.68 \sin 42.5^{\circ}$
$Area=68.68(0.6756)$
$Area=46.400=46.4$
Therefore, the area of the triangle is $46.4$ m$^{2}$.