You already know that triangle area can be found with $A=\frac{1}{2}\times base\times height$. In this sketch, $CQ=a$ is the base and $CP=b$.

         P
        /|
       / |
    b /  | h
     /   |
    /    |
   C-----D------Q
   <------ a ------>

Angle $C$ is between sides $a$ and $b$. Dropping a perpendicular from $P$ to the base at $D$ gives the height $PD=h$. In right triangle $CPD$, $\sin C=\frac{PD}{CP}=\frac{h}{b}$, so $h=b\sin C$. Substituting that into the usual area formula gives

$$A=\frac{1}{2}ah=\frac{1}{2}a(b\sin C)=\frac{1}{2}ab\sin C$$

So when two sides and their included angle are known, use $A=\frac{1}{2}ab\sin C$. More generally, for any two sides $x$ and $y$ with included angle $\theta$, $A=\frac{1}{2}xy\sin \theta$. The angle in the formula must be the one directly between the two sides you choose. Area answers should be given in squared units such as ${cm}^2$ or $m^2$.