中四math..點計？
2008-05-07 7:05 am
回答 (3)
2008-05-10 5:29 am
✔ 最佳答案
第１題 在∆ABD中,
∠BAC=180˚-45˚-90˚-30˚(∆內角和)
=15˚
∠ACB=90˚+45˚(∆外角)
=135˚
AB/sin135˚ =25/sin15
AB=25sin135˚ /sin15˚
x=ABsin30˚
=34.2cm(準確至３位有效數字)
第２題
cos∠ACB=(62+72-82)/[2(6)(7)]
∠ACB≈75.5225˚
∠ACD=180˚-75.5225˚(直線上的鄰角)
=104.4775˚
x2=62+52-2(6)(5)cos104.4775˚
x=8.72cm(準確至３位有效數字)
2008-05-08 12:34 am
1.)from the graph, angle CAD = 45 ,so CD = x , BD=25+x
x/(25+x)=tan30
x/(25+x)=0.577
14.425+0.577x=x
0.423X=14.425
x=34.1
2.)use cos law,
8^2=6^2+7^2-2*6*7*cosACB->(angle)
cosACB=0.25
ACE=75.522
ACD=180-75.522=104.478
x^2=5^2+6^2-2*5*6*cos104.478
x=8.72
x/(25+x)=tan30
x/(25+x)=0.577
14.425+0.577x=x
0.423X=14.425
x=34.1
2.)use cos law,
8^2=6^2+7^2-2*6*7*cosACB->(angle)
cosACB=0.25
ACE=75.522
ACD=180-75.522=104.478
x^2=5^2+6^2-2*5*6*cos104.478
x=8.72
2008-05-07 7:26 am
tan30=x/25+x
tan30(25)+tan30(x)=x
tan30(25)=x-tan30(x)
tan30(25)=x(1-tan30)
x=(tan30(25))/(1-tan30)
x=34.1506....(個人計算,不能盡信)
tan30(25)+tan30(x)=x
tan30(25)=x-tan30(x)
tan30(25)=x(1-tan30)
x=(tan30(25))/(1-tan30)
x=34.1506....(個人計算,不能盡信)
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