We have
We will use the Law of cosines in order to find the angle C
[tex]c^2=a^2+b^2-2ab\cdot cos\mleft(C\mright)[/tex]we need to isolate C
[tex]\cos (C)=\frac{c^2-a^2-b^2}{-2ab}[/tex]where
a=3 ft
b = 6 ft
c= 7 ft.
we substitute the values
[tex]\cos (C)=\frac{7^2-3^2-6^2}{-2(3)(6)}=-\frac{1}{9}[/tex][tex]C=\cos ^{-1}(-\frac{1}{9})=96.38\text{\degree}[/tex]Then we will use the law of sines
[tex]\frac{\sin(96.38)}{7}=\frac{\sin(A)}{3}[/tex][tex]\sin (A)=\frac{3\cdot\sin (96.38)}{7}[/tex][tex]\sin (A)=0.426[/tex][tex]A=\sin ^{-1}(0.426)=25.21[/tex]Then we use the fact that the sum of the interior angles of a triangle is 180°
A+B+C=180
B=180-25.21-96.38=58.41°
The solution is
A= 25.21 degrees
B= 58.41 degrees
C= 96.38 degrees
