trig

whats the angle for C if a = 8 b = 6 c = 7

cos-rule

$\begin{array}{rcl} c^2 &=& a^2+b^2-2ab\cos{(C)} \\ c^2+2ab\cos{(C)} &=& a^2+b^2 \\ 2ab\cos{(C)} &=& a^2+b^2-c^2 \\ \cos{(C)} &=& \frac{a^2+b^2-c^2} {2ab} \\ \cos{(C)} &=& \frac{8^2+6^2-7^2} {2\cdot 8 \cdot 6} \\ \cos{(C)} &=& \frac{64+36-49} {96} \\ \cos{(C)} &=& \frac{51} {96} = \frac{17} {32} \\ \cos{(C)} &=& 0.53125 \\ C &=& 57.9100487437^\circ \end{array}$

I am not sure, wanna keep me company?