字體：小 中 大 |
|
|
|
| 2017/07/04 06:08:21瀏覽41｜回應0｜推薦0 | |
| 標題: Applications in Trigonometry 發問: 圖片參考：http://imgcld.yimg.com/8/n/HA00967484/o/701106040077813873455670.jpg a(i)我識做.. Answers : (a)(ii) 35.8度 (b) P: 11.3度,Q: 21.8度
此文章來自奇摩知識+如有不便請留言告知 最佳解答:a ii) cos ∠KBC = BC/BD = 3/5 Then KC2 = 52 + 92 - 2 x 5 x 9 cos ∠KBC = 52 KC = √52 Thus tan θ = √52/10 θ = 35.8 b) tan ∠BPD = BD/DP = 3/2 ∠BPD = 56.3 With Φ = 45, P has to turn through an angle of 56.3 - 45 = 11.3 QK2 = QC2 + CK2 = 152QB2 = QC2 + CB2 = 181Then in triangle QKB:KB2 = QB2 + QK2 - 2(QB)(QK) cos ∠KQBcos ∠KQB = 0.928cos ∠KQB = 21.8 其他解答: |
|
| ( 時事評論｜政治 ) |
