1. **Problem statement:** Find the angle $\theta$ for point A(-3,4) in the domain $0 < \theta \leq 4\pi$. 2. **Formula and rules:** The angle $\theta$ in standard position is found using the arctangent function: $$\theta = \arctan\left(\frac{y}{x}\right)$$ However, since arctan only returns values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, we must adjust $\theta$ based on the quadrant. 3. **Determine the quadrant:** Point A(-3,4) lies in the second quadrant because $x<0$ and $y>0$. 4. **Calculate the reference angle:** $$\alpha = \arctan\left|\frac{4}{-3}\right| = \arctan\left(\frac{4}{3}\right)$$ Using a calculator: $$\alpha \approx 0.93$$ radians. 5. **Find $\theta$ in the second quadrant:** $$\theta = \pi - \alpha = 3.1416 - 0.93 = 2.21$$ radians. 6. **Check domain:** $0 < 2.21 \leq 4\pi$ is true, so $\theta = 2.21$ radians. **Final answer:** $$\boxed{\theta \approx 2.21 \text{ radians}}$$