三角比的關係

題目給了你的比,你可以用畢氏定理找出第三條邊的長度
那麼其他的三角比你也能找出了
多教一樣,就是,tanθ=sinθ/cosθ

有根號的粗體並斜體了

2. sinθ=0.25=1/4

so, the 3 sides are 1,
4^2-1^2 = 15 ,
4,

cosθ = 15 / 4, tanθ = 1/ 15

5. cosθ=7分之1

so, the 3 sides are 7^2-1^2 = 48 = 4(3)
1
7

sinθ = 4(3)/7, tanθ = 4(3)/1 = 4(3)

8. tanθ = 1.25 = 5/4

so, the 3 sides are 5
4
5^2+4^2 = 41

sinθ = 1/41 = 41/41 cosθ = 4/41 = 4(41)/41

11. tanθ = 8分之15

so, the 3 sides are 8
15
8^2+15^2 = 17

3sinθ + 5cosθ = 3(8/17) + 5(15/17) = 99/17

14. tanθ / sinθ = (sinθ/cosθ) / sinθ = 1 / cosθ = 1/0.2 = 5

sinθ=0.25=1/4

=>sin^2θ=1/16

=>1-sin^2θ=15/16

=>cos^2θ=15/16

=>cosθ=sqrt(15)=4

tanθ=sinθ/cosθ=(1/4)/(sqrt15/4)=1/sqrt15//

--------------------------

cosθ=1/7

=>cos^2θ=1/49

=>1-cso^2θ=48/49

=>sin^2θ=48/49

=>sinθ=4(sqrt3)/7

=>tanθ=sinθ/cosθ=(4sqrt3/7)/(1/7)=4sqrt(3)//

-------------------------

tanθ=1.25=5/4

since tanθ=sinθ/cosθ

then sinθ=5k and cosθ=4k , where k is a constant

sin^2θ+cos^2θ=1

=>(5k)^2+(4k)^2=1

=>41k^2=1

=>k=1/sqrt(41)

so sinθ=5/sqrt(41)0 and cosθ=4/sqrt(41)//

------------------------

11. tanθ=8/15 -> sinθ/cosθ=8/15

sinθ=8k and cosθ=15k , where k is a constant

sin^2θ+cos^2θ=1

=>64k^2+225k^2=1

=>289k^2=1

=>k=1/17

so sinθ=8/17 and cosθ=15/17//

hence 3sinθ+5cosθ=(8*3+15*5)/17=99/17//




2010-03-05 21:21:20 補充：
tanθ/sinθ

=(sinθ/cosθ)/sinθ

=1/cosθ

=1/(0.2)

=5//

2010-03-05 21:46:14 補充：
查一兩個字而已 , 又可以學英文 , 何樂而不為 ?

2010-03-05 23:17:16 補充：
螞蟻雄兵 , θ 是一隻銳角 , 0 < θ <90 呀!!