F.3 maths problem

Hi I am STY. My solutions are as follows:

Identities:

sin^2θ + cos^2θ = 1

tan(90 - θ) = 1 / tanθ

So,

sinθ = 1/4

cos^2θ = 1 - sin^2θ

= 1 - (1/4)^2

= 1 - 1/16

= 15/16

cosθ = sqrt15 / 4

tan(90 - θ) = 1 / tanθ

= 1 / (sinθ / cosθ)

= cosθ / sinθ

= (sqrt15 / 4) / (1 / 4)

= sqrt15

註：sqrt = 開方

有不明白的地方可再問我。

Plz feel free to ask if there are any extra questions.


BY STY~~~````

作直角三角形ABC,當中AB=4,AC=1,而AB,AC,BC分別代表斜邊,對邊和鄰邊
BC=開方(4^2-1^2)=開方15
cosθ=(開方15)/4
tanθ=1/開方15
tan(90度-θ)=1/tanθ=1/(1/開方15)=開方15

圖片參考：http://imgcld.yimg.com/8/n/HA00900668/o/701109010081113873452431.jpg


因為tan(90度-θ)=1/tanθ
所以tan(90庋-θ)=√15

第2幅圖入面錯左,Y果條邊係√15

有問題可以再問~

COS θ＝開方15 ／ 4
TAN θ＝開方15 ／15