f4 maths

2010-05-24 4:24 am

回答 (1)

2010-05-24 5:15 am
✔ 最佳答案

圖片參考:http://farm5.static.flickr.com/4013/4631307561_250da0173d_b.jpg



5) △OAC + sector OCB

= (1/2)(OA)(OC)sinㄥAOC + (TT r^2 )ㄥCOB/360°

= (1/2)(r^2)sin120° + (TT r^2)(60/360)

= (√3/4)r^2 + (TT r^2)/6

= (r^2)(√3/4 +TT/6)

So (r^2)(√3/4) + TT/6) = 25√3 + 50TT/3

r^2 = 100

r = 10cm




b)cosㄥACD = (5^2 + 15^2 - 12^2)/(2*5*15) = 106/150


ㄥACD = 45.03565°


cosㄥDAC = (12^2 + 15^2 - 5^2)/(2*12*15) = 344/360

ㄥDAC = 17.14621°

So ㄥDAB = 45.03565 + 17.14621 = 62.18186°

In △ADB ,by sine law ,

(sinㄥABD) / 12 = (sinㄥDAB) / 17

(sinㄥABD) / 12 = (sin62.18186°) / 17

sinㄥABD = 0.62431

ㄥABD = 38.6°(3 sig.fig.) = ㄥBDC



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