Solution.
We are required to find the area triangle to the nearest tenth.
Two sides of the triangle and an included angle is given
[tex]\begin{gathered} Thus,\text{ the area of the triangle is }\frac{1}{2}absin\theta \\ Where\text{ a,b= sides of the triangle} \\ \theta=included\text{ angle} \end{gathered}[/tex][tex]\begin{gathered} For\text{ the triangle given, } \\ a=15m \\ \text{ b=8m} \\ \theta=108 \end{gathered}[/tex][tex]Area=\frac{1}{2}\text{ x 15 x 8 x sin108}[/tex][tex]\begin{gathered} Area=57.0623m^2 \\ Area\text{ of the triangle = 57.1m}^2(nearest\text{ tenth} \end{gathered}[/tex]
