1. **State the problem:** We have quadrilateral ABCD with given angles $m\angle B = 65^\circ$, $m\angle D = 48^\circ$, and sides $CD = 15$, $AD = 21$, $AB = 17$. Points B, C, and D are not collinear. We want to find all possible values of $m\angle ACB$ to the nearest degree. 2. **Use the Law of Cosines to find $AC$ in triangle ABD:** Triangle ABD has sides $AB = 17$, $AD = 21$, and angle $m\angle B = 65^\circ$ between them. The Law of Cosines formula: $$AC^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(m\angle B)$$ Substitute values: $$AC^2 = 17^2 + 21^2 - 2 \cdot 17 \cdot 21 \cdot \cos(65^\circ)$$ Calculate: $$AC^2 = 289 + 441 - 714 \cdot \cos(65^\circ)$$ $$AC^2 = 730 - 714 \cdot 0.4226$$ $$AC^2 = 730 - 301.8 = 428.2$$ So, $$AC = \sqrt{428.2} \approx 20.7$$ 3. **Use the Law of Sines in triangle ACB to find $m\angle ACB$:** We know $AB = 17$, $AC \approx 20.7$, and $CD = 15$. Since B, C, D are not collinear and $BC + CD = BD$, but $BC$ is unknown, we consider triangle ACB. Using Law of Sines: $$\frac{\sin(m\angle ACB)}{AB} = \frac{\sin(m\angle BAC)}{BC} = \frac{\sin(m\angle ABC)}{AC}$$ We want $m\angle ACB$, so focus on: $$\sin(m\angle ACB) = \frac{AB}{AC} \sin(m\angle ABC)$$ Angle $m\angle ABC$ is given as $65^\circ$. Calculate: $$\sin(m\angle ACB) = \frac{17}{20.7} \times \sin(65^\circ) = 0.8217 \times 0.9063 = 0.7447$$ 4. **Find possible values of $m\angle ACB$:** Since sine is positive in the first and second quadrants, $$m\angle ACB = \sin^{-1}(0.7447) \approx 48^\circ$$ or $$m\angle ACB = 180^\circ - 48^\circ = 132^\circ$$ 5. **Final answer:** The possible values for $m\angle ACB$ are approximately $48^\circ$ and $132^\circ$.