Step 1: Understanding the Question:
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$.
Step 3: Detailed Explanation:
- Line 1: $4x - 2y + 13 = 0 \Rightarrow 2y = 4x + 13 \Rightarrow y = 2x + \frac{13}{2}$.
Slope $m_1 = 2$.
- Line 2: Makes equal intercepts with axes.
Let intercepts be $a$. Equation: $\frac{x}{a} + \frac{y}{a} = 1 \Rightarrow x + y = a \Rightarrow y = -x + a$.
Slope $m_2 = -1$. (Note: If intercepts were equal in magnitude but opposite sign, $m$ would be 1, but "equal intercepts" implies $a=b$).
- Calculating $\tan \theta$:
$\tan \theta = \left| \frac{2 - (-1)}{1 + 2(-1)} \right| = \left| \frac{3}{1 - 2} \right| = \left| \frac{3}{-1} \right| = 3$.
So, $\theta = \tan^{-1}(3)$.
Step 4: Final Answer:
The acute angle is $\tan^{-1}(3)$.