在△ABC中,已知A=π/4,cosB=2ㄏ5/5.求:（1）求sinC的值?（2）若BC=2ㄏ5,D为AB的中点,求C

1）∵A=π/4,sinA=√2/2,cosA=√2/2
∵cosB=2√5/5,∴sinB=√5/5
∴sinC=sin(180°-(A+B))
=sin(A+B)
=sinA*cosB+cosA*sinB
=√2/2×2√5/5+√2/2×√5/5=3√10/10
2）正弦定理得：sinA/BC=sinC/AB,
AB=(3√10/10*2√5)/√2/2=6,∴BD=3
余弦定理得：CD²=BD²+BC²-2cosB*BD*BC
=9+20-2*3*2√5*2√5/5=5
∴CD=√5

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