急!F5 Solving Trinages 10q4

Bid

2013-01-28 6:05 am

請詳細步驟教我計以下二條 :
不要網址回答:

圖片參考：http://imgcld.yimg.com/8/n/HA05788109/o/20130127220132.jpg

第29的A及B我識做,只是唔識C,而30條就完全唔識做
29
a) <CPA=30,度 <BCP=15度
b) PC=2x, BC=x開方2 , BP=(開方3-1)x

回答 (1)

myisland8132

2013-01-28 6:23 am

✔ 最佳答案

29(a) sin∠CPA = AC/PC = 1/2 =>∠CPA = 30∠BCP = 45 - 30 = 15(b) PC = 2AC = 2xBC^2 = AC^2 + AB^2 = 2x^2 => BC = √2xAP^2 + AC^2 = PC^2AP^2 + x^2 = 4x^2So, AP = √3x => BP = (√3 - 1)xcos∠BCP = (PC^2 + BC^2 - BP^2)/2(PC)(BC)= (4 + 2 - 4 + 2√3)/(2 * 2 * √2)= (1 + √3)/(2 * √2)= (√6 + √2)/430(a) Area of ABCD= Area BCD + Area ABD= (1/2)(BC)(BD)sinθ + (1/2)(AB)(BD)sin(90 - θ)= (1/2)(BD)(BCsinθ + ABcosθ)(b)(i) AC = 10 cmsinθ = CE/BC = BC/AC = 3/5cosθ = √(1 - 9/25) = 4/5(ii) Area of ABCD= (1/2)(BD)(BCsinθ + ABcosθ)= (1/2)(10)(6 * 3/5 + 8 * 4/5)= 18 + 32= 50 cm^2

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