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(10) 1B. Suppose have a triangle with a 120 degree angle opposite a missing side. If the other two sides are 6 inches and 7 inches, find the third side and the two other angles if possible. Round to the nearest hundredth. Is it possible? Yes or No I used the law of cosines, which states c^2 = a^2 + b^2 - 2ab(cos C) and the law of sines, which states sinA/a = sinB/b = sinC/c. The side opposite the 120 degree angle is 11.27 inches. The angle opposite 6 inches is 27.46 degrees. The angle opposite 7 inches is 32.54 degrees. Is there a possibility of more than one triangle that fits this description? Yes or No State why there is or is not the possibility of more than one answer. The value for C is fixed at C = 120 degrees. I don't have any extra range to put in another triangle. This is a angle-side-side therefore, I will use the law of cosines. What do I have? Side a and b. Solve for c. c^2 = a^2 + b^2 - 2ab(cos C) A: 27.46 a: 6 in. B: 32.54 b: 7 in. C: 120 c: 27.47 solve for c c^2 = 6^2 + 7^2 - 2(6)(7)(cos 120) c^2 = 36 + 49 - 2(42)(cos 120) c^2 = 85 - 84(cos 120) c^2 = 85 - 84(-1/2) c^2 = 85 + 42 c^2 = 127 c = sqrt(127) c approx 11.26942767 solve for A sin 120 / 11.27 = sin A / 6 (6 sin 120) / 11.27 = sin A sin^-1((6 sin 120) / 11.27) = A A approx 27.45556419 27.46 solve for B sin 120 / 11.27 = sin B / 7 in. (7) sin 120 / 11.27 = sin B B = sin^-1((7 sin 120) / 11.27) B approx 32.54106719

          (10) 1B. Suppose have a triangle with a 120 degree angle opposite a missing side. If the other two sides are 6 inches and 7 inches, find the third side and the two other angles if possible. Round to the nearest hundredth. Is it possible? Yes or No I used the law of cosines, which states c^2 = a^2 + b^2 - 2ab(cos C) and the law of sines, which states sinA/a = sinB/b = sinC/c. The side opposite the 120 degree angle is 11.27 inches. The angle opposite 6 inches is 27.46 degrees. The angle opposite 7 inches is 32.54 degrees. Is there a possibility of more than one triangle that fits this description? Yes or No State why there is or is not the possibility of more than one answer. The value for C is fixed at C = 120 degrees. I don't have any extra range to put in another triangle. This is a angle-side-side therefore, I will use the law of cosines. What do I have? Side a and b. Solve for c. c^2 = a^2 + b^2 - 2ab(cos C) A: 27.46 a: 6 in. B: 32.54 b: 7 in. C: 120 c: 27.47 solve for c c^2 = 6^2 + 7^2 - 2(6)(7)(cos 120) c^2 = 36 + 49 - 2(42)(cos 120) c^2 = 85 - 84(cos 120) c^2 = 85 - 84(-1/2) c^2 = 85 + 42 c^2 = 127 c = sqrt(127) c approx 11.26942767 solve for A sin 120 / 11.27 = sin A / 6 (6 sin 120) / 11.27 = sin A sin^-1((6 sin 120) / 11.27) = A A approx 27.45556419 27.46 solve for B sin 120 / 11.27 = sin B / 7 in. (7) sin 120 / 11.27 = sin B B = sin^-1((7 sin 120) / 11.27) B approx 32.54106719

(10) 1B. Suppose have a triangle with a 120 degree angle opposite a missing side. If the other two sides are 6 inches and 7 inches, find the third side and the two other angles if possible. Round to the nearest hundredth. Is it possible? Yes or No I used the law of cosines, which states c^2 = a^2 + b^2 - 2ab(cos C) and the law of sines, which states sinA/a = sinB/b = sinC/c. The side opposite the 120 degree angle is 11.27 inches. The angle opposite 6 inches is 27.46 degrees. The angle opposite 7 inches is 32.54 degrees. Is there a possibility of more than one triangle that fits this description? Yes or No State why there is or is not the possibility of more than one answer. The value for C is fixed at C = 120 degrees. I don't have any extra range to put in another triangle. This is a angle-side-side therefore, I will use the law of cosines. What do I have? Side a and b. Solve for c. c^2 = a^2 + b^2 - 2ab(cos C) A: 27.46 a: 6 in. B: 32.54 b: 7 in. C: 120 c: 27.47 solve for c c^2 = 6^2 + 7^2 - 2(6)(7)(cos 120) c^2 = 36 + 49 - 2(42)(cos 120) c^2 = 85 - 84(cos 120) c^2 = 85 - 84(-1/2) c^2 = 85 + 42 c^2 = 127 c = sqrt(127) c approx 11.26942767 solve for A sin 120 / 11.27 = sin A / 6 (6 sin 120) / 11.27 = sin A sin^-1((6 sin 120) / 11.27) = A A approx 27.45556419 27.46 solve for B sin 120 / 11.27 = sin B / 7 in. (7) sin 120 / 11.27 = sin B B = sin^-1((7 sin 120) / 11.27) B approx 32.54106719

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