中四附加數(急!!)

2009-03-11 6:42 am
1). 圖中OAB是一個半徑扇形,已知該扇形的周界是(20+4π)cm,求(http://x00.xanga.com/3ddf13f737031236005740/s186436445.png)
a). 求 r 的值
b). 求陰影的面積
c). 求陰影的周界
2). 化簡 a). [tan(90∘+θ)cos(θ+270∘)sin(90∘+θ)] /
[cot(180∘-θ)sin(θ+270∘)]
b).(http://x91.xanga.com/504f03f643630236008144/s186438591.png)
3). 証明下列恆等式 :
(tanθsinθ)/(tanθ-sinθ)≡(tanθ+sinθ)/(tanθsinθ)
4).解下列方程 :
3cos^2θ-sinθcosθ=1,0≤ θ ≥360∘

回答 (2)

2009-03-12 9:02 am
✔ 最佳答案
1.
a)設角為θ
扇形周界就係:
r+r+rθ=20+4π

2r+rθ=20+4π

孤度問題多數r都係整數,而度數多數係π既陪數

所以:

2r=20

r=10

rθ=4π

10θ=4π

θ=0.4π rad. or θ=72度

b)陰影的面積

=扇形面積-三角形面積

=10(4π)-0.5*sin72 10^2

=78.1 cm^2 (取至3個有效數字)

c) 個三角形剩底個條邊既邊長^2

=10^2+10^2-2*10*10*cos 72

=200-200cos72

邊長=開方上面既答案

周界=邊長+4π

=24.3cm (取至3個有效數字)

2009-03-12 01:23:55 補充:
2)

a)[tan(90+θ)cos(θ+270)sin(90+θ)] /[cot(180-θ)sin(θ+270)]

=(cotθ sinθ cos θ)/[cotθ cos(-θ)]

(約去cotθ cosθ)

=-sinθ

2009-03-12 01:24:36 補充:
b)tan(π+θ)/[sin(π+θ) cos(3π-θ)] +[sin(4π-θ)sin(π+θ)]/{sin[(3π/2)+θ] cos(-θ-6π)}

=tan(180+θ)/[sin(180+θ) cos(540-θ)]+[sin(720-θ)sin(180+θ)/[sin(270+θ)cos-(θ+1080)]

=tanθ/sin-θ cos-θ+ (sin-θ sin-θ)/(cos-θ cosθ)

=tanθ/sinθ cosθ+ sin^2 θ/-cos^2 θ

2009-03-12 01:24:44 補充:
=(sinθ/cosθ)/sinθcosθ-tan^2 θ

=1/cos^2 θ-tan^2 θ

=sec^2 θ-tan^2 θ

=(tan^2 θ+1)-tan^2 θ

=1
2009-03-12 1:05 am
2,3 )
http://hk.knowledge.yahoo.com/question/question?qid=7008073103391

4) The required angle = 45, 117, 225 or 297

2009-03-11 17:06:14 補充:
1) The angle is not given, so it is not possible to calculate.
參考: My knowledge & Yahoo+ knowledge


收錄日期: 2021-04-23 19:12:02
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