2. \( * * \) SHOW WORK ON PAPER REQUIRED FOR THIS QUESTION*** A pool ball bounces off the sides of a rectangular pool table in such a way that \( \angle 1 \cong \angle 3, \angle 4 \cong \angle 6 \), and \( \angle 3 \) and \( \angle 4 \) are complementary. If \( m \angle 1=33^{\circ} \), find \( m \angle 3, m \angle 4 \), and \( m \angle 5 \). \( m \angle 3= \) \( m \angle 4= \) \( m \angle 5= \)

Certainly! Let's work through the problem step by step. ### **Given:** 1. \( \angle 1 \cong \angle 3 \) 2. \( \angle 4 \cong \angle 6 \) 3. \( \angle 3 \) and \( \angle 4 \) are **complementary** (i.e., \( \angle 3 + \angle 4 = 90^\circ \)) 4. \( m \angle 1 = 33^\circ \) ### **Find:** - \( m \angle 3 \) - \( m \angle 4 \) - \( m \angle 5 \) ### **Solution:** 1. **Determine \( m \angle 3 \):** - Given that \( \angle 1 \cong \angle 3 \) and \( m \angle 1 = 33^\circ \), it follows that: \[ m \angle 3 = m \angle 1 = 33^\circ \] 2. **Determine \( m \angle 4 \):** - Since \( \angle 3 \) and \( \angle 4 \) are complementary: \[ m \angle 3 + m \angle 4 = 90^\circ \] Substitute \( m \angle 3 = 33^\circ \) into the equation: \[ 33^\circ + m \angle 4 = 90^\circ \] Solve for \( m \angle 4 \): \[ m \angle 4 = 90^\circ - 33^\circ = 57^\circ \] 3. **Determine \( m \angle 5 \):** - Without a diagram, we'll make a common geometric assumption: if \( \angle 4 \) and \( \angle 5 \) form a linear pair (they are supplementary), then: \[ m \angle 4 + m \angle 5 = 180^\circ \] Substitute \( m \angle 4 = 57^\circ \) into the equation: \[ 57^\circ + m \angle 5 = 180^\circ \] Solve for \( m \angle 5 \): \[ m \angle 5 = 180^\circ - 57^\circ = 123^\circ \] ### **Answer:** \[ \begin{align*} m \angle 3 &= 33^\circ \\ m \angle 4 &= 57^\circ \\ m \angle 5 &= 123^\circ \\ \end{align*} \]