問關於三角學

tanθ = 2
(2sinθ + 3cosθ) / (3sinθ - 2cosθ)
= [2tanθ + 3 ]/ [3tanθ - 2]
= [2(2) + 3] / [3(2) - 2]
= 7/5


圖片參考：http://i54.photobucket.com/albums/g111/Tim-Mx/clip_image002.jpg

a)
x = ytan75
b)
x = (y + 50)tan55
c)
by(a) and(b)
ytan75 = (y + 50)tan55
ytan75 = ytan55 + 50tan55
y(tan75 - tan55) = 50tan55
y = 50tan55/(tan75 - tan55)
y = 40.0 m(3 sig.fig)
x = 116 m(3 sig.fig)

2007-04-04 22:45:08 補充：
sorry for the mistakes of yy = 30.0 m(3.sig.fig)

1.since tanθ=2,use the idenity tanθ=sinθ/cosθ

so,sinθ/cosθ=2
sinθ=2cosθ
subst sinθ=2cosθ,
4cosθ+3cosθ/6cosθ-2cosθ
=7cosθ/4cosθ
=7/4

2a. tan 75=x/y
b. tan 55=x/(y+50)
c.from 2a,
x=y tan 75
from 2b,
x=(y+50) tan 55
so,y tan75=(y+50) tan 55
y(tan 75-tan 55)=50 tan 55
y=31.0(correct to 3 sig. fig.)
subst y=31.0,
tan 75=x/31.0
x=116(correct to 3 sig. fig.)
so,x=116,y=31.0