1. **State the problem:** Find the acute angle $\theta$ between two lines $a$ and $b$. Line $a$ has equation $x - 2y + 1 = 0$. Line $b$ makes an angle of $60^\circ$ with the positive x-axis. 2. **Find the slope of line $a$:** Rewrite $a$ in slope-intercept form $y = mx + c$: $$x - 2y + 1 = 0 \implies -2y = -x - 1 \implies y = \frac{1}{2}x + \frac{1}{2}$$ So, slope of $a$ is $m_1 = \frac{1}{2}$. 3. **Find the slope of line $b$:** Line $b$ makes an angle of $60^\circ$ with the positive x-axis, so its slope is: $$m_2 = \tan 60^\circ = \sqrt{3}$$ 4. **Formula for angle between two lines:** The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ 5. **Calculate $\tan \theta$:** $$\tan \theta = \left| \frac{\frac{1}{2} - \sqrt{3}}{1 + \frac{1}{2} \times \sqrt{3}} \right| = \left| \frac{\frac{1}{2} - \sqrt{3}}{1 + \frac{\sqrt{3}}{2}} \right|$$ 6. **Simplify numerator and denominator:** Numerator: $$\frac{1}{2} - \sqrt{3} = \frac{1 - 2\sqrt{3}}{2}$$ Denominator: $$1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$$ 7. **Divide numerator by denominator:** $$\tan \theta = \left| \frac{\frac{1 - 2\sqrt{3}}{2}}{\frac{2 + \sqrt{3}}{2}} \right| = \left| \frac{1 - 2\sqrt{3}}{2} \times \frac{2}{2 + \sqrt{3}} \right| = \left| \frac{1 - 2\sqrt{3}}{2 + \sqrt{3}} \right|$$ 8. **Rationalize denominator:** Multiply numerator and denominator by the conjugate $2 - \sqrt{3}$: $$\tan \theta = \left| \frac{(1 - 2\sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \right|$$ Denominator: $$(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$$ Numerator: $$(1)(2) - (1)(\sqrt{3}) - (2\sqrt{3})(2) + (2\sqrt{3})(\sqrt{3}) = 2 - \sqrt{3} - 4\sqrt{3} + 6 = 8 - 5\sqrt{3}$$ 9. **So:** $$\tan \theta = |8 - 5\sqrt{3}|$$ 10. **Calculate numeric value:** $$\sqrt{3} \approx 1.732$$ $$8 - 5 \times 1.732 = 8 - 8.66 = -0.66$$ Absolute value: $$| -0.66 | = 0.66$$ 11. **Find $\theta$:** $$\theta = \arctan(0.66) \approx 33.690^\circ$$ 12. **Final answer:** The acute angle between lines $a$ and $b$ is approximately **33.690 degrees** (to 3 decimal places).