law of cosine

law of cosine

a=6 b=9 c=10 find the angles

$\begin{array}{rcll} \hline a^2 &=& b^2+c^2-2bc\cdot \cos{(A)} \\ 2bc\cdot \cos{(A)} &=& b^2+c^2-a^2 \\ \cos{(A)} &=& \frac{ b^2+c^2-a^2 } {2bc} \\ \cos{(A)} &=& \frac{ 9^2+10^2-6^2 } {2\cdot 9 \cdot 10} \\ \cos{(A)} &=& \frac{ 81+100-36 } {180} \\ \cos{(A)} &=& \frac{ 145 } {180} \\ \cos{(A)} &=& 0.80555555556 \\ A &=& \arccos{(0.80555555556)} \\ A &=& 36.3360575146^{\circ} \\\\ \hline b^2 &=& c^2+a^2-2ca\cdot \cos{(B)} \\ 2ca\cdot \cos{(B)} &=& c^2+a^2-b^2 \\ \cos{(B)} &=& \frac{ c^2+a^2-b^2 } {2ca} \\ \cos{(B)} &=& \frac{ 10^2+6^2-9^2 } {2\cdot 10 \cdot 6} \\ \cos{(B)} &=& \frac{ 100+36-81 } {120} \\ \cos{(B)} &=& \frac{ 55 } {120} \\ \cos{(B)} &=& 0.45833333333 \\ B &=& \arccos{(0.45833333333)} \\ B &=& 62.7203872640^{\circ}\\\\ \hline c^2 &=& a^2+b^2-2ab\cdot \cos{(C)} \\ 2ab\cdot \cos{(C)} &=& a^2+b^2-c^2 \\ \cos{(C)} &=& \frac{ a^2+b^2-c^2 } {2ab} \\ \cos{(C)} &=& \frac{ 6^2+9^2-10^2 } {2\cdot 6\cdot 9} \\ \cos{(C)} &=& \frac{ 36+81-100 } {108} \\ \cos{(C)} &=& \frac{ 17 } {108} \\ \cos{(C)} &=& 0.15740740741 \\ C &=& \arccos{(0.15740740741)} \\ C &=& 80.9435552214^{\circ} \\ \hline \end{array}$

Which is oppistie which is adjacent and which is the hyptounuse

Mark them like this O for oppisite A for adjacent and H for hypotonuse.