area、simplify、change subject、ratio、pentagon area 20+20分 = 40

(99) Angle CBD =1/2 AngleCOD (Angle at centre= 2 Angle at circumference)
Angle CBD= 21
Angle BCO=Angle COD=42 (alt Angle BC//OD)
x= Angle CBD + Angle OCB =21+42 =63 (ext Angle of triangle)

(101) OC=2, OA=4, so AC=6
Then find the intersection of BC and AB
Put x=4 into x-2y+2=0
4-2y+2=0
-2y=-6
y=3
so B=(4,3), AB=3
Area of Triangle ABC= 1/2(6)(3)=9 sq units

(103) 先處理 x-1/x, 其他唔寫住, x-1/x= (x^2-1)/x
所以 1-2x/(x-1/x) =1-2x/((x^2-1)/x)
=1-2x^2/x^2-1
=(x^2-1-2x^2)/x^2-1 (通分母)
=-(x^2+1)/(x^2-1)

(104) a=2-1/1+b = 2(1+b)-1/1+b = 1+2b/1+b
a(1+b)=1+2b=a+ab
b(a-2) = 1-a
b= 1-a/a-2

(106) Let AB=3k, BC=4k where k is a constant
畢氏定理, AC=5k
sinθ = BC/AC= 4k/5k =4/5

(107) 這個正五邊形可以分成五個相同的等腰三角形, 中間的角度由360度分成5份, 每份中間角為72度, 所以其餘的兩隻角= (180-72)/2 = 54
利用sine law, 設x是等腰三角形的其餘兩邊
x / sin54 = 2 / sin72 , x=2sin54/sin72

Area of regular pentagon= 1/2(2x) sin 54 (5) = 6.88 cm^2

2007-09-01 02:55:07 補充：
(107) 6.88cm^2 (nearest 0.01cm^2)