標題:
有a.maths 唔識做
發問:
最佳解答:
1. sinA=5/13, cosA=12/13 cosB=4/5, sinB=3/5 cosC = cos(180-A-B) =-cos(A+B) =-[cosAcosB-sinAsinB] =-[(12/13)(4/5)-(5/13)(3/5)] =-33/65 2. sinα = 9/41, cosα = 40/41 cos(α+β) = 7/25 , sin(α+β) = 24/25 sinβ = sin(α+ β- α) = sin(α+ β) cos(α) - cos(α+ β) sin(α) =(24/25)(40/41) - (7/25)(9/41) =(960-63)/1025 =(960-63)/1025 =897/1025
其他解答:
有a.maths 唔識做
發問:
此文章來自奇摩知識+如有不便請留言告知
1) In△ABC, sinA = 5/13, cos B = 4/5 , find the value of cos C 2) If sinα = 9/41, cos(α+β) = 7/25 and α, βare acute angle , find the value of sinβ最佳解答:
1. sinA=5/13, cosA=12/13 cosB=4/5, sinB=3/5 cosC = cos(180-A-B) =-cos(A+B) =-[cosAcosB-sinAsinB] =-[(12/13)(4/5)-(5/13)(3/5)] =-33/65 2. sinα = 9/41, cosα = 40/41 cos(α+β) = 7/25 , sin(α+β) = 24/25 sinβ = sin(α+ β- α) = sin(α+ β) cos(α) - cos(α+ β) sin(α) =(24/25)(40/41) - (7/25)(9/41) =(960-63)/1025 =(960-63)/1025 =897/1025
其他解答:
文章標籤
全站熱搜
留言列表
