1. **Problem statement:** Find the quadrant in which the rotating line OP lies for the angle $\frac{7\pi}{6}$ and find the acute angle that OP makes with the x-axis. 2. **Formula and rules:** - The angle $\theta = \frac{7\pi}{6}$ radians. - One full rotation is $2\pi$ radians. - Quadrants are divided as: - Quadrant I: $0 < \theta < \frac{\pi}{2}$ - Quadrant II: $\frac{\pi}{2} < \theta < \pi$ - Quadrant III: $\pi < \theta < \frac{3\pi}{2}$ - Quadrant IV: $\frac{3\pi}{2} < \theta < 2\pi$ - The acute angle with the x-axis is the smallest positive angle between OP and the x-axis. 3. **Determine the quadrant:** - Since $\pi = \frac{6\pi}{6}$ and $\frac{3\pi}{2} = \frac{9\pi}{6}$, - $\frac{7\pi}{6}$ lies between $\pi$ and $\frac{3\pi}{2}$, - So, OP lies in **Quadrant III**. 4. **Find the acute angle with the x-axis:** - The reference angle $\alpha = \theta - \pi = \frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}$. - $\frac{\pi}{6}$ radians is the acute angle OP makes with the x-axis. 5. **Summary:** - The line OP lies in Quadrant III. - The acute angle with the x-axis is $\frac{\pi}{6}$ radians or 30°. **Final answer:** $$\text{Quadrant} = \text{III}, \quad \text{Acute angle} = \frac{\pi}{6}$$