Help Geometry ASAP

In triangle ABC, sin A : sin B : sin C = 5 : 5 : 6. Find cos C.

$\begin{array}{|lrcll|} \hline (1): & \mathbf{\dfrac{\sin(A)}{\sin(B)}} &=& \mathbf{\dfrac{5}{5}} \\\\ & \dfrac{\sin(A)}{\sin(B)} &=& 1 \\ \\ & \sin(A) &=& \sin(B) \\ & \mathbf{A} &=& \mathbf{B} \quad | \quad A=180^\circ-B \text{ is not possible}\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \sin(C) &=& \sin\Big(180^\circ-(A+B)\Big) \quad | \quad B=A \\ \sin(C) &=& \sin\Big(180^\circ-(A+A)\Big) \\ \sin(C) &=& \sin\Big(180^\circ-(2A)\Big) \\ \sin(C) &=& \sin( 2A ) \\ \mathbf{\sin(C)} &=& \mathbf{2\sin( A )\cos(A)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline (2): & \mathbf{\dfrac{\sin(A)}{\sin(C)}} &=& \mathbf{\dfrac{5}{6}} \\\\ & \dfrac{\sin(A)}{2\sin( A )\cos(A)} &=& \dfrac{5}{6} \\\\ & \dfrac{1}{2 \cos(A)} &=& \dfrac{5}{6} \\\\ & 2 \cos(A) &=& \dfrac{6}{5} \\\\ & \mathbf{ \cos(A) } &=& \mathbf{ \dfrac{6}{10} } \\ \hline & \sin(A) &=& \sqrt{1-\cos^2(A) } \\ & \sin(A) &=& \sqrt{1-\dfrac{6^2}{10^2} } \\ & \sin(A) &=& \sqrt{\dfrac{10^2-6^2}{10^2} } \\ & \sin(A) &=& \sqrt{\dfrac{8^2}{10^2} } \\ & \mathbf{ \sin(A) } &=& \mathbf{ \dfrac{8}{10} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline (2): & \mathbf{\dfrac{\sin(A)}{\sin(C)}} &=& \mathbf{\dfrac{5}{6}} \quad | \quad \mathbf{ \sin(A) = \dfrac{8}{10} } \\\\ & \dfrac{\dfrac{8}{10}}{\sin(C)} &=& \dfrac{5}{6} \\\\ & \sin(C) &=& \dfrac{8}{10}\cdot \dfrac{6}{5} \\\\ & \mathbf{ \sin(C) } &=& \mathbf{ \dfrac{24}{25} } \\ \hline & \cos(C) &=& \sqrt{1-\sin^2(C) } \\ & \cos(C) &=& \sqrt{1-\dfrac{24^2}{25^2} } \\ & \cos(C) &=& \sqrt{\dfrac{25^2-24^2}{25^2} } \\ & \cos(C) &=& \sqrt{\dfrac{7^2}{25^2} } \\ & \cos(C) &=& \dfrac{7 }{25 } \\ & \mathbf{\cos(C)} &=& \mathbf{0.28} \\ \hline \end{array}$

cos c ~ .279