Find the angle between lines AB and AC given points A(0, 5), B(-3, 0), C(4, 0).

Points: A(0, 5), B(-3, 0), C(4, 0).
Slope of AB ($m_{AB}$):

$$m_{AB} = \frac{0 - 5}{-3 - 0} = \frac{-5}{-3} = \frac{5}{3}$$

Slope of AC ($m_{AC}$):

$$m_{AC} = \frac{0 - 5}{4 - 0} = \frac{-5}{4}$$

Let $m_1 = m_{AC} = -\frac{5}{4}$ and $m_2 = m_{AB} = \frac{5}{3}$.

$$\tan \theta = \left| \frac{\frac{5}{3} - \left(-\frac{5}{4}\right)}{1 + \left(\frac{5}{3}\right)\left(-\frac{5}{4}\right)} \right|$$ $$\tan \theta = \left| \frac{\frac{20}{12} + \frac{15}{12}}{1 - \frac{25}{12}} \right|$$ $$\tan \theta = \left| \frac{\frac{35}{12}}{\frac{12 - 25}{12}} \right|$$ $$\tan \theta = \left| \frac{\frac{35}{12}}{-\frac{13}{12}} \right| = \left| -\frac{35}{13} \right| = \frac{35}{13}$$

$$\theta = \arctan\left(\frac{35}{13}\right) \approx 69.5906^\circ$$

Rounded to two decimal places, $\theta$ is 69.59°.

$$\boxed{\text{a) } \theta \approx 69.59^\circ}$$

Points: A(-1, 6), B(5, -1), C(-2, 0), D(3, 6).
Slope of AC ($m_{AC}$):

$$m_{AC} = \frac{0 - 6}{-2 - (-1)} = \frac{-6}{-1} = 6$$

Slope of BD ($m_{BD}$):

$$m_{BD} = \frac{6 - (-1)}{3 - 5} = \frac{7}{-2} = -\frac{7}{2}$$

Let $m_1 = m_{AC} = 6$ and $m_2 = m_{BD} = -\frac{7}{2}$.

$$\tan \theta = \left| \frac{-\frac{7}{2} - 6}{1 + (6)\left(-\frac{7}{2}\right)} \right|$$ $$\tan \theta = \left| \frac{-\frac{7}{2} - \frac{12}{2}}{1 - 21} \right|$$ $$\tan \theta = \left| \frac{-\frac{19}{2}}{-20} \right| = \left| \frac{19}{40} \right| = \frac{19}{40}$$

$$\theta = \arctan\left(\frac{19}{40}\right) \approx 25.3909^\circ$$

Rounded to two decimal places, $\theta$ is 25.39°.

$$\boxed{\text{b) } \theta \approx 25.39^\circ}$$

Points: A(-2, -1), C(0, 4), D(3, 2).
Slope of AC ($m_{AC}$):

$$m_{AC} = \frac{4 - (-1)}{0 - (-2)} = \frac{5}{2}$$

Slope of CD ($m_{CD}$):

$$m_{CD} = \frac{2 - 4}{3 - 0} = \frac{-2}{3}$$

Let $m_1 = m_{AC} = \frac{5}{2}$ and $m_2 = m_{CD} = -\frac{2}{3}$.

$$\tan \theta = \left| \frac{-\frac{2}{3} - \frac{5}{2}}{1 + \left(\frac{5}{2}\right)\left(-\frac{2}{3}\right)} \right|$$ $$\tan \theta = \left| \frac{-\frac{4}{6} - \frac{15}{6}}{1 - \frac{10}{6}} \right|$$ $$\tan \theta = \left| \frac{-\frac{19}{6}}{\frac{6 - 10}{6}} \right|$$ $$\tan \theta = \left| \frac{-\frac{19}{6}}{-\frac{4}{6}} \right| = \left| \frac{19}{4} \right| = \frac{19}{4}$$

$$\theta = \arctan\left(\frac{19}{4}\right) \approx 78.1109^\circ$$

Rounded to two decimal places, $\theta$ is 78.11°.

$$\boxed{\text{c) } \theta \approx 78.11^\circ}$$

Points: A(0, -6), B(0, 2), C(5, 0).
Slope of AC ($m_{AC}$):

$$m_{AC} = \frac{0 - (-6)}{5 - 0} = \frac{6}{5}$$

Slope of BC ($m_{BC}$):

$$m_{BC} = \frac{0 - 2}{5 - 0} = \frac{-2}{5}$$

Let $m_1 = m_{AC} = \frac{6}{5}$ and $m_2 = m_{BC} = -\frac{2}{5}$.

$$\tan \theta = \left| \frac{-\frac{2}{5} - \frac{6}{5}}{1 + \left(\frac{6}{5}\right)\left(-\frac{2}{5}\right)} \right|$$ $$\tan \theta = \left| \frac{-\frac{8}{5}}{1 - \frac{12}{25}} \right|$$ $$\tan \theta = \left| \frac{-\frac{8}{5}}{\frac{25 - 12}{25}} \right|$$ $$\tan \theta = \left| \frac{-\frac{8}{5}}{\frac{13}{25}} \right| = \left| -\frac{8}{5} \times \frac{25}{13} \right| = \left| -\frac{8 \times 5}{13} \right| = \left| -\frac{40}{13} \right| = \frac{40}{13}$$

$$\theta = \arctan\left(\frac{40}{13}\right) \approx 72.0046^\circ$$

Rounded to two decimal places, $\theta$ is 72.00°.

$$\boxed{\text{d) } \theta \approx 72.00^\circ}$$