Answer
$\displaystyle \cos A=\frac{4}{5}$
$\displaystyle \tan A=\frac{3}{4}$
Work Step by Step
In quadrant I, cosine is positive.
($\cos\theta=\pm\sqrt{1-\sin^{2}\theta}$ is with a plus sign)
$\displaystyle \cos A=+\sqrt{1-(\frac{3}{5})^{2}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
$\displaystyle \tan A=\frac{\sin A}{\cos A}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
