The given Triangle has sides
[tex] AB=17in, BC=8 in, m<C=58 [/tex]
using sine rule we can write
[tex] \frac{17}{sin58}=\frac{8}{sinA}\\\\sinA=\frac{8}{17}*sin58\\\\sinA= 0.39908\\\\m<A=23.52\\\\m<B=180-58-23.52=98.48\\\\ [/tex]
Area of Triangle =[tex] \frac{1}{2}absin \theta\\ [/tex]
[tex] A=\frac{1}{2}absin \theta \\\\A=\frac{1}{2}*17*8*sin(98.48)\\\\A=67.2565 \ in^2 \ \\\\A=67.3 \ in^2 \\ [/tex]
