數學知識交流---三角(3)

(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