Answer:
![\boxed{A=43.54cm^2}](https://img.qammunity.org/2020/formulas/mathematics/high-school/tspcj0lvze24k0i14xnwhlnfy8dgbzsdb5.png)
Explanation:
To find this area we will use the law of cosine and the Heron's formula. First of all, let't find the unknown side using the law of cosine:
![x^2=12^2+8^2-2(12)(8)cos(65^(\circ)) \\ \\ x^2=144+64-192(0.42) \\ \\ x^2=208-80.64 \\ \\ x^2=127.36 \\ \\ x=√(127.36) \\ \\ \therefore \boxed{x=11.28cm}](https://img.qammunity.org/2020/formulas/mathematics/high-school/q1qo5rvhw3w3pglny9vii69lh0t6ordeqt.png)
Heron's formula (also called hero's formula) is used to find the area of a triangle using the triangle's side lengths and the semiperimeter. A polygon's semiperimeter s is half its perimeter. So the area of a triangle can be found by:
being
the corresponding sides of the triangle.
So the semiperimeter is:
![s=(12+8+11.28)/(2) \\ \\ s=15.64cm](https://img.qammunity.org/2020/formulas/mathematics/high-school/mwuttat9ol0ysx925op4rtwm2fw78eu5qv.png)
So the area is:
![A=√(15.64(15.64-12)(15.64-8)(15.64-11.28)) \\ \\ \therefore \boxed{A=43.54cm^2}](https://img.qammunity.org/2020/formulas/mathematics/high-school/oh2ezw59kdpudrb117un87jos01vofdrwz.png)