在三角形ABC中,已知A=45度,cosB=4/5 求sinC . 2、若BC=10,D为AB中点,求CD长

cosA=sinA=√2/2
cosB=4/5,sinB=3/5
cosC=(180-45-B)
=sin(B-45)
=sinBcos45-sin45cosB
=-√2/10
sinC=7√2/10
AC/sinB=AB/sinC=BC/sinA=√2
AC=4√2/5,AB=7/5
AD=7/10
CD=√[(7/10)^2+(4√2/5)^2-2(7/10)(4√2/5)*cos45]
=11/10
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