(c) If \( 5 \cos \mathrm{~A}+3=0 \) and \( 180^{\circ}<\mathrm{A}<360^{\circ} \), determine by means of a diagram: \( \begin{array}{ll}\text { (1) } \tan ^{2} \mathrm{~A} & \text { (2) } \frac{\sin \mathrm{A}}{\cos \mathrm{A}}\end{array} \)

First, let’s solve for \( \cos A \). Given the equation \( 5 \cos A + 3 = 0 \), we can rearrange it to find: \[ \cos A = -\frac{3}{5} \] Since we are looking for angle \( A \) in the range \( 180^\circ < A < 360^\circ \), we know \( A \) is in the third or fourth quadrant where cosine is negative. Now, we can use the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \) to find \( \sin A \): \[ \sin^2 A = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \] Thus, \( \sin A = -\frac{4}{5} \) (since \( A \) must also be in the third quadrant where sine is negative). Now, let's proceed to calculate \( \tan^2 A \): (1) \[ \tan A = \frac{\sin A}{\cos A} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} \] So, \[ \tan^2 A = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \] (2) For \( \frac{\sin A}{\cos A} \), we already calculated that: \[ \frac{\sin A}{\cos A} = \tan A = \frac{4}{3} \] So, the final answers are: (1) \( \tan^2 A = \frac{16}{9} \) (2) \( \frac{\sin A}{\cos A} = \frac{4}{3} \)