We will have the following:
1.
[tex]\begin{gathered} A=(10)(12)\frac{sin(55)}{2}\Rightarrow A=60sin(55) \\ \\ \Rightarrow A\approx49.1 \end{gathered}[/tex]
So, the area of the first triangle is approximately 49.1 ft^2.
2.
[tex]\begin{gathered} A=(18.5)(25)\frac{sin(20)}{2}\Rightarrow A=231.25sin(20) \\ \\ \Rightarrow A\approx79.1 \end{gathered}[/tex]
So, the area of the second triangle is approximately 79.1 in^2.
