1. **Problem Statement:** Find the angle between the two lines given by the equations: $$x + 7y - 3 = 0$$ and $$x - y + 5 = 0$$ 2. **Formula to use:** The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ 3. **Find the slopes of the lines:** For the first line $x + 7y - 3 = 0$, rewrite in slope-intercept form $y = mx + c$: $$7y = -x + 3$$ $$y = \frac{-1}{7}x + \frac{3}{7}$$ So, $m_1 = -\frac{1}{7}$. For the second line $x - y + 5 = 0$: $$-y = -x - 5$$ $$y = x + 5$$ So, $m_2 = 1$. 4. **Calculate $\tan \theta$:** $$\tan \theta = \left| \frac{-\frac{1}{7} - 1}{1 + \left(-\frac{1}{7}\right)(1)} \right| = \left| \frac{-\frac{1}{7} - \frac{7}{7}}{1 - \frac{1}{7}} \right| = \left| \frac{-\frac{8}{7}}{\frac{6}{7}} \right|$$ 5. **Simplify the fraction:** $$\tan \theta = \left| \frac{\cancel{\frac{8}{7}}}{\cancel{\frac{6}{7}}} \times \frac{7}{7} \right| = \left| -\frac{8}{6} \right| = \frac{8}{6} = \frac{4}{3}$$ 6. **Find the angle $\theta$:** $$\theta = \arctan \left( \frac{4}{3} \right)$$ Using a calculator or known values: $$\theta \approx 53.13^\circ$$ **Final answer:** The angle between the two lines is approximately $53.13^\circ$.