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问题: 初三-圆

⊿ABC中,BC为直径的⊙0交AB于D,交AC于E,AD=3,S⊿ADE=S四边形BCED,CE=(1/2)√2(根号)
求∠A
求sinA
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解答:

过C作CF⊥AB于F
BC为直径,CD⊥AB
S⊿ADE=1/2AD*EF,S⊿ABC=1/2AB*CD
S⊿ADE=S四边形BCED,2S⊿ADE=S⊿ABC
2AD*EF=AB*CD

CD⊥AB,EF⊥AB,CD∥EF
EF/CD=AE/AC,EF=AE*CD/AC
2AD*AE*CD/AC=AB*CD
2AD*AE=AB*AC.....................(1)
又:
AD*AB=AE*AC......................(2)
(1)*(2)
2AD^2=AC^2,AC=3√2
CD=√(AC^2-AD^2)=3
sinA=CD/AC=3/(3√2)=√2/2
A=45°