Consider that the given diagram is a triangle ABC with base BC 13cm, side AC 7cm, and an included angle of 38 degrees.
Consider the following diagram,
Construction: Draw a perpendicular from A to BC at point M, such that AM represents the height of the triangle ABC.
Apply the sine ratio in the triangle AMC,
[tex]\begin{gathered} \sin \theta=\frac{\text{ Opposite Side}}{\text{ Hypotenuse}} \\ \sin 38^{\circ}=\frac{AM}{AC} \\ 0.61566=\frac{AM}{7} \\ AM=0.61566\times7 \\ AM\approx4.31 \end{gathered}[/tex]
Now, the area of the triangle is calculated as,
[tex]\begin{gathered} Area=\frac{1}{2}\times BC\times AM \\ Area=\frac{1}{2}\times13\times4.31 \\ Area\approx28.0 \end{gathered}[/tex]
Thus, the area of the given triangle is 28.0 square centimeters.
