Answer
$\displaystyle \cos A=-\frac{12}{13}$
$\displaystyle \tan A=\frac{12}{5}$
Work Step by Step
In quadrant III, sine is negative.
($\sin B=\pm\sqrt{1-\cos^{2}B}$ is with a minus sign)
$\sin B=-\sqrt{1-(-\frac{5}{13})^{2}}=\sqrt{\frac{169-25}{169}}$
$=-\displaystyle \sqrt{\frac{144}{169}}=-\frac{12}{13}$
$\displaystyle \tan A=\frac{\sin A}{\cos A}=\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}$
