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

We are given \[ 5\cos A + 3 = 0. \] Solving for \(\cos A\): \[ 5\cos A = -3 \quad \Longrightarrow \quad \cos A = -\frac{3}{5}. \] Since \(180^\circ < A < 360^\circ\), the angle lies in either the third or fourth quadrant. The cosine is negative, which means \(A\) must be in the **third quadrant** where both sine and cosine are negative. Next, we use the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1. \] Substitute \(\cos A = -\frac{3}{5}\): \[ \sin^2 A + \left(-\frac{3}{5}\right)^2 = 1 \quad \Longrightarrow \quad \sin^2 A + \frac{9}{25} = 1. \] Solve for \(\sin^2 A\): \[ \sin^2 A = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}. \] Since \(A\) is in the third quadrant, \(\sin A\) is negative: \[ \sin A = -\frac{4}{5}. \] Now, we can determine the following: 1. \(\tan^2 A\) Recall that \[ \tan A = \frac{\sin A}{\cos A} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}. \] Thus, \[ \tan^2 A = \left(\frac{4}{3}\right)^2 = \frac{16}{9}. \] 2. \(\frac{\sin A}{\cos A}\) This is just the tangent of \(A\): \[ \frac{\sin A}{\cos A} = \tan A = \frac{4}{3}. \] To help visualize, consider the following diagram description in the third quadrant: - Draw a coordinate plane. - Plot a point corresponding to angle \(A\) in the third quadrant. - In a right triangle formed by dropping a perpendicular from the point to the \(x\)-axis, let the adjacent side (along the \(x\)-axis) have length \(3\) (taking absolute value) and the opposite side (vertical) have length \(4\) (taking absolute value). Since the point is in the third quadrant, both coordinates are negative, so you have: - Adjacent side (negative \(x\)-direction): \(-3\) - Opposite side (negative \(y\)-direction): \(-4\) Using these, the hypotenuse is: \[ r = \sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5. \] Then: \[ \cos A = \frac{-3}{5}, \quad \sin A = \frac{-4}{5}, \quad \tan A = \frac{-4/5}{-3/5}=\frac{4}{3}. \] Thus, the required answers are: \[ \tan^2 A = \frac{16}{9} \quad \text{and} \quad \frac{\sin A}{\cos A} = \frac{4}{3}. \]