在三角形ABC中,∠A,∠B,∠C所对边的长分别为a,b,c,

∵cos∠ABC=√6/6
∴sin∠ABC=√30/6
在△BCD中,BD/sinC=DC/sin∠DBC
DC*sinC=BD*sin∠DBC
在△ABC中,AC/sin∠ABC=AB/sinC
AC*sinC=AB*sin∠ABC
又DC=(1/2)*AC
∴BD*sin∠DBC=(1/2)*AB*sin∠ABC
则 sin∠DBC=(1/2)*AB*sin∠ABC/BD
=(1/2)×(√6/3)×(√30/6)÷√5
=2/3
∴ cos∠DBC=√5/3
故 sin∠ABD=sin(∠ABC-∠DBC)
=sin∠ABC*cos∠DBC-cos∠ABC*sin∠DBC
=(√30/6)×(√5/3)-(√6/6)×(2/3)
=√6/6
∴cos∠ABD=√30/6
在△ABD中,cos∠ABD=(AB²+BD²-AD²)/2AB*BD
∴ AD²=AB²+BD²-2AB*BD*cos∠ABD
=(4√6/3)²+(√5)²-2×(4√6/3)×(√5)×(√30/6)
=7/3
∴ AD=√21/3
∵ BD/sinA=AD/sin∠ABD
∴ sinA=BD*sin∠ABD/AD
=(√5)×(√6/6)÷(√21/3)
=√70/14