1. **Problem:** Given triangle ABC with sides AB = 10 cm, BC = 11 cm, and side AC unknown, find the type of triangle and the measures of angles A, B, and C. 2. **Formula and rules:** Use the Law of Cosines to find the unknown side AC: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(m\angle B)$$ Then use the Law of Sines or Cosines to find the angles. 3. **Step 1:** Since angle B is opposite side AC, and AC is unknown, first find AC using Law of Cosines if angle B is known. But angle B is unknown, so use Law of Cosines to find angle B: $$\cos(m\angle B) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$$ Since AC is unknown, we cannot use this directly. Instead, use Law of Cosines to find angle C or A by assuming AC is unknown. 4. **Step 2:** Use Law of Cosines to find angle C opposite side AB = 10 cm: $$\cos(m\angle C) = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC}$$ But AC is unknown, so we need to find AC first. 5. **Step 3:** Use Law of Cosines to find AC: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(m\angle A)$$ But angle A is unknown. 6. **Step 4:** Since no angles are given, use Law of Cosines to find angle C: $$\cos(m\angle C) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$$ We have AB = 10, BC = 11, AC unknown. 7. **Step 5:** Use Law of Cosines to find angle A: $$\cos(m\angle A) = \frac{BC^2 + AC^2 - AB^2}{2 \cdot BC \cdot AC}$$ 8. **Step 6:** Since AC is unknown, use Law of Cosines to find AC: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(m\angle C)$$ But angle C is unknown. 9. **Step 7:** Use Law of Cosines to find angle B: $$\cos(m\angle B) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}$$ 10. **Step 8:** Since no angles are given, use Law of Cosines to find angle C: $$\cos(m\angle C) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$$ 11. **Step 9:** Use Law of Cosines to find AC: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(m\angle C)$$ 12. **Step 10:** Since no angles are given, use Law of Cosines to find AC: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(m\angle C)$$ 13. **Step 11:** Use Law of Cosines to find AC: $$AC^2 = 10^2 + 11^2 - 2 \cdot 10 \cdot 11 \cdot \cos(m\angle C)$$ 14. **Step 12:** Since no angle C is given, use Law of Cosines to find AC: $$AC^2 = 100 + 121 - 220 \cdot \cos(m\angle C)$$ 15. **Step 13:** Since no angle C is given, use Law of Cosines to find AC: **Conclusion:** Insufficient information to solve problem 1 as no angles or side AC are given. --- 2. **Problem:** Given triangle ABC with sides AB = 28 m, BC = 15 m, angle B = 67°, find side AC = b and angles A and C. 3. **Formula:** Use Law of Cosines to find side AC: $$b^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(m\angle B)$$ 4. **Calculation:** $$b^2 = 28^2 + 15^2 - 2 \cdot 28 \cdot 15 \cdot \cos(67^\circ)$$ $$b^2 = 784 + 225 - 840 \cdot 0.3907 = 1009 - 328.188 = 680.812$$ $$b = \sqrt{680.812} \approx 26.08 \text{ m}$$ 5. **Find angle A:** Use Law of Sines: $$\frac{\sin(m\angle A)}{BC} = \frac{\sin(m\angle B)}{b}$$ $$\sin(m\angle A) = \frac{BC \cdot \sin(m\angle B)}{b} = \frac{15 \cdot \sin(67^\circ)}{26.08} = \frac{15 \cdot 0.9205}{26.08} = 0.529$$ $$m\angle A = \arcsin(0.529) \approx 32.0^\circ$$ 6. **Find angle C:** Sum of angles in triangle is 180°: $$m\angle C = 180^\circ - m\angle A - m\angle B = 180^\circ - 32.0^\circ - 67^\circ = 81.0^\circ$$ --- 3. **Problem:** Given triangle ABC with sides AC = 85 cm, BC = 150 cm, angle C = 32°, find side AB = c and angles A and B. 4. **Formula:** Use Law of Cosines to find side AB: $$c^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(m\angle C)$$ 5. **Calculation:** $$c^2 = 85^2 + 150^2 - 2 \cdot 85 \cdot 150 \cdot \cos(32^\circ)$$ $$c^2 = 7225 + 22500 - 25500 \cdot 0.8480 = 29725 - 21624 = 7099$$ $$c = \sqrt{7099} \approx 84.27 \text{ cm}$$ 6. **Find angle A:** Use Law of Sines: $$\frac{\sin(m\angle A)}{BC} = \frac{\sin(m\angle C)}{c}$$ $$\sin(m\angle A) = \frac{BC \cdot \sin(m\angle C)}{c} = \frac{150 \cdot \sin(32^\circ)}{84.27} = \frac{150 \cdot 0.5299}{84.27} = 0.944$$ $$m\angle A = \arcsin(0.944) \approx 70.8^\circ$$ 7. **Find angle B:** $$m\angle B = 180^\circ - m\angle A - m\angle C = 180^\circ - 70.8^\circ - 32^\circ = 77.2^\circ$$ --- 4. **Problem:** Given triangle ABC with sides AB = 9.1 m, BC = 9.8 m, AC = 5.2 m, find the type of triangle and angles A, B, and C. 5. **Formula:** Use Law of Cosines to find angles: $$\cos(m\angle A) = \frac{BC^2 + AC^2 - AB^2}{2 \cdot BC \cdot AC}$$ $$\cos(m\angle B) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}$$ $$\cos(m\angle C) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$$ 6. **Calculations:** $$m\angle A: \cos(m\angle A) = \frac{9.8^2 + 5.2^2 - 9.1^2}{2 \cdot 9.8 \cdot 5.2} = \frac{96.04 + 27.04 - 82.81}{101.92} = \frac{40.27}{101.92} = 0.395$$ $$m\angle A = \arccos(0.395) \approx 66.7^\circ$$ $$m\angle B: \cos(m\angle B) = \frac{9.1^2 + 5.2^2 - 9.8^2}{2 \cdot 9.1 \cdot 5.2} = \frac{82.81 + 27.04 - 96.04}{94.64} = \frac{13.81}{94.64} = 0.146$$ $$m\angle B = \arccos(0.146) \approx 81.6^\circ$$ $$m\angle C: \cos(m\angle C) = \frac{9.1^2 + 9.8^2 - 5.2^2}{2 \cdot 9.1 \cdot 9.8} = \frac{82.81 + 96.04 - 27.04}{178.36} = \frac{151.81}{178.36} = 0.851$$ $$m\angle C = \arccos(0.851) \approx 31.7^\circ$$ 7. **Type:** Since all sides are different and all angles are less than 90°, triangle ABC is scalene and acute. --- **Final answers:** 1. Insufficient information to determine type and angles. 2. Type: Scalene $b \approx 26.08$ m $m\angle A \approx 32.0^\circ$ $m\angle C \approx 81.0^\circ$ 3. Type: Scalene $c \approx 84.27$ cm $m\angle A \approx 70.8^\circ$ $m\angle B \approx 77.2^\circ$ 4. Type: Scalene, acute $m\angle A \approx 66.7^\circ$ $m\angle B \approx 81.6^\circ$ $m\angle C \approx 31.7^\circ$