標題:
數學知識交流---三角(3)
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA01076848/o/701108280069213873450240.jpg 求 AB , AC , BC , ∠A , ∠B , ∠C 若 (1) BC = 67 , AC = 41 , ∠A = 118° (2) BC = 4 , AC = 2 , ∠B = 27° (3) BC = 29 , AC = 17 , ∠C = 41°
最佳解答:
(1) BC = 67 , AC = 41 , ∠A = 118° 解： I. 由正弦定律得： AC / sin B = BC / sin A sin B = AC × sin A / BC sin B = 41 × sin 118° / 67 sin B ≈ 41 × 0.8829 / 67 sin B ≈ 0.5403 ∠B ≈ 32.7° II. 由三角形內角和定理得： ∠C = 180° - ∠A - ∠B = 180° - 118° - 32.7° = 29.3° III. 由正弦定律得： BC / sin A = AB / sin C AB = BC × sin C / sin A AB = 67 × sin 29.3° / sin 118° AB ≈ 67 × 0.4894 / 0.8829 AB ≈ 37.1 (2) BC = 4 , AC = 2 , ∠B = 27° 解： I. 由正弦定律得： AC / sin B = BC / sin A sin A = BC × sin B / AC sin A = 4 × sin 27° / 2 sin A ≈ 4 × 0.4540 / 2 sin A = 0.908 ∠A ≈ 65.2° II. 由三角形內角和定理得： ∠C = 180° - ∠A - ∠B = 180° - 65.2° - 27° = 87.8° III. 由正弦定律得： BC / sin A = AB / sin C AB = BC × sin C / sin A AB = 4 × sin 87.8° / sin 65.2° AB ≈ 4 × 0.9993 / 0.4195 AB ≈ 9.5 (3) BC = 29 , AC = 17 , ∠C = 41° 解： I. 由餘弦定律得： AB2 = BC2 + AC2 - 2 × BC × AC × cos C AB2 = 292 + 172 - 2 × 29 × 17 × cos 41° AB2 ≈ 1130 - 986 × 0.7547 AB2 = 385.8658 AB = 19.6 II. (∠A + ∠B) / 2 = 90° - ∠C / 2 (∠A + ∠B) / 2 = 69.5° -----(1) 由正切定律得： (BC - AB) / (BC + AB) = [tan (∠A-∠B)/2] / [tan (∠A+∠B)/2] (29 - 19.6) / (29 + 19.6) = [tan (∠A-∠B)/2] / tan 69.5° 0.1934 × tan 69.5° = tan (∠A-∠B)/2 0.1934 × 2.6746 = tan (∠A-∠B)/2 tan (∠A-∠B)/2 = 0.5173 (∠A - ∠B) / 2 ≈ 27.4° ---------(2) (1) + (2)，得： ∠A = 96.9° (1) - (2)，得： ∠B = 42.1° 2011-08-28 19:27:57 補充： (1) AB = 37.135413848295642515387678068367791754767926034065658121212 BC = 67 AC = 41 ∠A = 118° ∠B = 32.70482701260594219729943890307588391150738639444858834366° ∠C = 29.29517298739405780270056109692411608849261360555141165634° 驗算： (BC - AC) / AB = [sin (A-B) / 2] / [sin (A+B) / 2] 2011-08-28 19:38:25 補充： (2) AB = 4.40204878042008022183246118099112778904470077755931077046 BC = 4 AC = 2 ∠A = 65.2278149068694621964181226550684873670267091740164977945° ∠B = 27° ∠C = 87.7721850931305378035818773449315126329732908259835022055° 2011-08-28 19:39:12 補充： 看意見。 2011-08-28 19:40:25 補充： 更正： AB = BC × sin C / sin A AB = 4 × sin 87.8° / sin 65.2° AB ≈ 4 × 0.9993 / 0.9078 AB ≈ 4.4 驗算： (BC - AC) / AB = [sin (A-B) / 2] / [sin (A+B) / 2] 2011-08-28 21:12:14 補充： 3. 更正： 由正切定律得： (BC - AC) / (BC + AC) = [tan (∠A-∠B)/2] / [tan (∠A+∠B)/2] (29 - 17) / (29 + 17) = [tan (∠A-∠B)/2] / tan 69.5° 0.2609 × tan 69.5° = tan (∠A-∠B)/2 0.2609 × 2.6746 = tan (∠A-∠B)/2 tan (∠A-∠B)/2 = 0.6978 (∠A - ∠B) / 2 ≈ 34.9° ---------(2) (1) + (2)，得： ∠A = 104.4° (1) - (2)，得： ∠B = 34.6° 2011-08-28 21:22:57 補充： (3) AB = 19.643226667234352565778561923564644079737787962992944144387 BC = 29 AC = 17 ∠A = 104.40453527002684937860089694901264102529749572607590888383° ∠B = 34.59546472997315062139910305098735897470250427392409111617° ∠C = 41° 2011-08-28 21:23:15 補充： 驗算：BC / sin A = AC / sin B = AB / sin C
數學知識交流---三角(3)
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA01076848/o/701108280069213873450240.jpg 求 AB , AC , BC , ∠A , ∠B , ∠C 若 (1) BC = 67 , AC = 41 , ∠A = 118° (2) BC = 4 , AC = 2 , ∠B = 27° (3) BC = 29 , AC = 17 , ∠C = 41°
最佳解答:
(1) BC = 67 , AC = 41 , ∠A = 118° 解： I. 由正弦定律得： AC / sin B = BC / sin A sin B = AC × sin A / BC sin B = 41 × sin 118° / 67 sin B ≈ 41 × 0.8829 / 67 sin B ≈ 0.5403 ∠B ≈ 32.7° II. 由三角形內角和定理得： ∠C = 180° - ∠A - ∠B = 180° - 118° - 32.7° = 29.3° III. 由正弦定律得： BC / sin A = AB / sin C AB = BC × sin C / sin A AB = 67 × sin 29.3° / sin 118° AB ≈ 67 × 0.4894 / 0.8829 AB ≈ 37.1 (2) BC = 4 , AC = 2 , ∠B = 27° 解： I. 由正弦定律得： AC / sin B = BC / sin A sin A = BC × sin B / AC sin A = 4 × sin 27° / 2 sin A ≈ 4 × 0.4540 / 2 sin A = 0.908 ∠A ≈ 65.2° II. 由三角形內角和定理得： ∠C = 180° - ∠A - ∠B = 180° - 65.2° - 27° = 87.8° III. 由正弦定律得： BC / sin A = AB / sin C AB = BC × sin C / sin A AB = 4 × sin 87.8° / sin 65.2° AB ≈ 4 × 0.9993 / 0.4195 AB ≈ 9.5 (3) BC = 29 , AC = 17 , ∠C = 41° 解： I. 由餘弦定律得： AB2 = BC2 + AC2 - 2 × BC × AC × cos C AB2 = 292 + 172 - 2 × 29 × 17 × cos 41° AB2 ≈ 1130 - 986 × 0.7547 AB2 = 385.8658 AB = 19.6 II. (∠A + ∠B) / 2 = 90° - ∠C / 2 (∠A + ∠B) / 2 = 69.5° -----(1) 由正切定律得： (BC - AB) / (BC + AB) = [tan (∠A-∠B)/2] / [tan (∠A+∠B)/2] (29 - 19.6) / (29 + 19.6) = [tan (∠A-∠B)/2] / tan 69.5° 0.1934 × tan 69.5° = tan (∠A-∠B)/2 0.1934 × 2.6746 = tan (∠A-∠B)/2 tan (∠A-∠B)/2 = 0.5173 (∠A - ∠B) / 2 ≈ 27.4° ---------(2) (1) + (2)，得： ∠A = 96.9° (1) - (2)，得： ∠B = 42.1° 2011-08-28 19:27:57 補充： (1) AB = 37.135413848295642515387678068367791754767926034065658121212 BC = 67 AC = 41 ∠A = 118° ∠B = 32.70482701260594219729943890307588391150738639444858834366° ∠C = 29.29517298739405780270056109692411608849261360555141165634° 驗算： (BC - AC) / AB = [sin (A-B) / 2] / [sin (A+B) / 2] 2011-08-28 19:38:25 補充： (2) AB = 4.40204878042008022183246118099112778904470077755931077046 BC = 4 AC = 2 ∠A = 65.2278149068694621964181226550684873670267091740164977945° ∠B = 27° ∠C = 87.7721850931305378035818773449315126329732908259835022055° 2011-08-28 19:39:12 補充： 看意見。 2011-08-28 19:40:25 補充： 更正： AB = BC × sin C / sin A AB = 4 × sin 87.8° / sin 65.2° AB ≈ 4 × 0.9993 / 0.9078 AB ≈ 4.4 驗算： (BC - AC) / AB = [sin (A-B) / 2] / [sin (A+B) / 2] 2011-08-28 21:12:14 補充： 3. 更正： 由正切定律得： (BC - AC) / (BC + AC) = [tan (∠A-∠B)/2] / [tan (∠A+∠B)/2] (29 - 17) / (29 + 17) = [tan (∠A-∠B)/2] / tan 69.5° 0.2609 × tan 69.5° = tan (∠A-∠B)/2 0.2609 × 2.6746 = tan (∠A-∠B)/2 tan (∠A-∠B)/2 = 0.6978 (∠A - ∠B) / 2 ≈ 34.9° ---------(2) (1) + (2)，得： ∠A = 104.4° (1) - (2)，得： ∠B = 34.6° 2011-08-28 21:22:57 補充： (3) AB = 19.643226667234352565778561923564644079737787962992944144387 BC = 29 AC = 17 ∠A = 104.40453527002684937860089694901264102529749572607590888383° ∠B = 34.59546472997315062139910305098735897470250427392409111617° ∠C = 41° 2011-08-28 21:23:15 補充： 驗算：BC / sin A = AC / sin B = AB / sin C
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