(1)根据正弦定理a/sinA=b/sinB,所以sin(A-π/3)=sinA,根据0<A<π解得A=2π/3
(2)在CD上取一点E,使得DE=2。这是AD=BD=ED=2,所以A、B、E在圆D上,且BE为直径,所以EA⊥AB
作AB边上的高CF, 则EA//CF,从而可以观察到等比例线段
设AF=x,因为∠CAF=π/3,所以AC=2x,CF=√3x
根据等比例线段,AB=4x
在∆CBF中应用勾股定理
3x^2+25x^2=25
x^2=25/28
所以S∆ABC=AB×CF/2=2√3x^2=25√3/14
S∆ADC=3/5×S∆ABC=15√3/14
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