Trigonometry problem, please help?
回答 (3)
2016-03-01 8:36 am
Right angled triangle with sides 3 , 4 , 5
cos ø = 3/5
sin ø = 4/5
cos ø = 3/5
sin ø = 4/5
2016-03-01 7:22 am
secϕ = 5/3
1/cosϕ = 5/3
cosϕ = 3/5
1- cos²ϕ = 1 - (3/5)²
sin²ϕ = 16/25
sinϕ = √(16/25) or sinϕ = -√(16/25)
sinϕ = 4/5 or sinϕ = -4/5
1/cosϕ = 5/3
cosϕ = 3/5
1- cos²ϕ = 1 - (3/5)²
sin²ϕ = 16/25
sinϕ = √(16/25) or sinϕ = -√(16/25)
sinϕ = 4/5 or sinϕ = -4/5
2016-03-01 10:49 am
sec(ø)=5/3 so
cos(ø)=3/5.
Since sin^2ø=1-cos^2ø = 1-9/25=16/25,
then sin(ø) =-4/5 or +4/5
cos(ø)=3/5.
Since sin^2ø=1-cos^2ø = 1-9/25=16/25,
then sin(ø) =-4/5 or +4/5
收錄日期: 2021-04-18 14:32:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160229231041AAHeu4n