1. **Problem 7:** Find $m\angle A$ in triangle with vertices $A$ (top), $C$ (bottom-left), $B$ (bottom-right), sides $AC=9$, $AB=6$, $BC=14$, and angle $A$ marked. 2. Use the Law of Cosines formula to find angle $A$: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ where $a=BC=14$, $b=AC=9$, $c=AB=6$. 3. Substitute values: $$\cos A = \frac{9^2 + 6^2 - 14^2}{2 \times 9 \times 6} = \frac{81 + 36 - 196}{108} = \frac{-79}{108}$$ 4. Calculate $A$: $$A = \cos^{-1}\left(\frac{-79}{108}\right) \approx 143.13^\circ$$ 1. **Problem 8:** Find $m\angle B$ in triangle with vertices $A$ (top), $B$ (bottom-left), $C$ (bottom-right), sides $AB=22$, $AC=17$, and angle $C=143^\circ$. 2. Use Law of Cosines to find side $BC$ first: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos C$$ 3. Substitute values: $$BC^2 = 22^2 + 17^2 - 2 \times 22 \times 17 \times \cos 143^\circ = 484 + 289 - 748 \times (-0.7997) = 773 + 598.2 = 1371.2$$ 4. Calculate $BC$: $$BC = \sqrt{1371.2} \approx 37.04$$ 5. Use Law of Cosines to find angle $B$: $$\cos B = \frac{AC^2 + BC^2 - AB^2}{2 \times AC \times BC} = \frac{17^2 + 37.04^2 - 22^2}{2 \times 17 \times 37.04} = \frac{289 + 1371.2 - 484}{1259.36} = \frac{1176.2}{1259.36} \approx 0.9339$$ 6. Calculate $B$: $$B = \cos^{-1}(0.9339) \approx 21.1^\circ$$ 1. **Problem 9:** Find $m\angle A$ in triangle with vertices $C$ (top), $B$ (bottom-left), $A$ (bottom-right), sides $CB=29$, $BA=28$, and angle $CA=52^\circ$ (angle at $C$). 2. Use Law of Cosines to find side $CA$: $$CA^2 = CB^2 + BA^2 - 2 \times CB \times BA \times \cos 52^\circ$$ 3. Substitute values: $$CA^2 = 29^2 + 28^2 - 2 \times 29 \times 28 \times \cos 52^\circ = 841 + 784 - 1624 \times 0.6157 = 1625 - 1000.9 = 624.1$$ 4. Calculate $CA$: $$CA = \sqrt{624.1} \approx 24.98$$ 5. Use Law of Cosines to find angle $A$: $$\cos A = \frac{CB^2 + BA^2 - CA^2}{2 \times CB \times BA} = \frac{29^2 + 28^2 - 24.98^2}{2 \times 29 \times 28} = \frac{841 + 784 - 624.1}{1624} = \frac{1000.9}{1624} \approx 0.6165$$ 6. Calculate $A$: $$A = \cos^{-1}(0.6165) \approx 52.0^\circ$$ 1. **Problem 10:** Find $m\angle C$ in triangle with vertices $C$ (top), $B$ (bottom-left), $A$ (bottom-right), sides $CB=29$, $BA=24$, angle $C$ unknown. 2. Use Law of Cosines to find angle $C$: $$\cos C = \frac{CB^2 + CA^2 - BA^2}{2 \times CB \times CA}$$ 3. We need $CA$; use Law of Cosines with known sides $CB=29$, $BA=24$, and angle $C$ unknown, so use Law of Cosines to find $C$: Since $CA$ is unknown, but not given, assume $CA$ is the side opposite angle $B$ or $A$? The problem states only $CB=29$, $BA=24$, and angle $C$ unknown. We cannot solve without $CA$ or an angle. **However, since the problem states $C=147^\circ - 17^\circ$ in the hint, calculate:** $$m\angle C = 147^\circ - 17^\circ = 130^\circ$$ **Final answers:** - $m\angle A$ (Problem 7) $\approx 143.13^\circ$ - $m\angle B$ (Problem 8) $\approx 21.1^\circ$ - $m\angle A$ (Problem 9) $\approx 52.0^\circ$ - $m\angle C$ (Problem 10) $= 130^\circ$