在△ABC中,已知b^2-bc-2c^2=0,且a=根号6,cosA=7/8,求S△ABC?

b²-bc-2c²=0,
=> (b-2c)(b+c)=0,
=> b=2c或b=-c,
b>0,c>0,
=> b=2c,
cosA=7/8,
=> sinA=√(1-cos²A)=√15/8,
cosA=(b²+c²-a²)/(2bc)=(5c²-6)/(4c²)=7/8,
=> 7×4c²=8×(5c²-6),
=> c=2,
=> b=4,
=> S△ABC=bcsinA/2=4×2×√15/8/2=√15/2
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