數學知識交流---三角(4)

若題目給出三角形的三邊，方法如下：
餘弦定律：
cos A = ( AC² + AB² - BC² ) / (2×AC×AB)
cos B = ( AB² + BC² - AB² ) / (2×AB×BC)
cos C = ( BC² + AC² - AB² ) / (2×BC×AC)
驗算：∠A + ∠B + ∠C = 180°

(1) AB = 3 , AC = 3 , BC = 5
解：
cos A = (3² + 3² - 5²) / (2 × 3 × 3) = -7/18
∠A ≈ 112.8853805°
cos B = (3² + 5² - 3²) / (2 × 3 × 5) = 5/6
∠B ≈ 33.55730976°
cos C = (5² + 3² - 3²) / (2 × 5 × 3) = 5/6
∠C ≈ 33.55730976°

(2) AB = 14 , AC = 13 , BC = 20
解：
cos A = (13² + 14² - 20²) / (2 × 13 × 14) = -5/52
∠A ≈ 95.51773437°
cos B = (14² + 20² - 13²) / (2 × 14 × 20) = 61/80
∠B ≈ 40.31490918°
cos C = (20² + 13² - 14²) / (2 × 20 × 13) = 373/520
∠C ≈ 44.16735645°

(3) AB = 29 , AC = 27 , BC = 50
解：
cos A = (27² + 29² - 50²) / (2 × 27 × 29) = -155/261
∠A ≈ 126.4320997°
cos B = (29² + 50² - 27²) / (2 × 29 × 50) = 653/725
∠B ≈ 25.75113208°
cos C = (50² + 27² - 29²) / (2 × 50 × 27) = 199/225
∠C ≈ 27.81676821°

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1）3^2=3^2+5^2-2*3*5cos ∠C
3^2=3^2+5^2-2*3*5cos ∠C
180-∠C-∠B= ∠A

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