3sinθ-2cosθ
——————
2cosθ
PS:若tanθ=4
le題數點計(maths)
2007-08-24 1:30 pm
回答 (2)
2007-08-24 5:12 pm
✔ 最佳答案
3sinθ-2cosθ——————
2cosθ
=(3sinθ / 2cosθ) - (2cosθ / 2cosθ)
=(3/2)*tanθ - 1
=(3/2)*4 - 1
=6 - 1
=5
2007-08-24 2:15 pm
For a right-angle triangle, angle A is 90deg, side a, b, c have relationship of
a^2 = b^2 + c^2
tan(B)= b/c = 4
sin(B)= b/a
cos(B)= c/a
(3sinB - 2cosB)/(2cosB)=(3*b/a -2*c/a)/(2*c/a)
=(3b-2c)/2c
=1.5*(b/c)-1
=1.5*4-1
=5
a^2 = b^2 + c^2
tan(B)= b/c = 4
sin(B)= b/a
cos(B)= c/a
(3sinB - 2cosB)/(2cosB)=(3*b/a -2*c/a)/(2*c/a)
=(3b-2c)/2c
=1.5*(b/c)-1
=1.5*4-1
=5
收錄日期: 2021-04-13 13:05:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070824000051KK00671