?$解:(1)證明:過(guò)點(diǎn)A作AD⊥BC,垂足為點(diǎn)D,如圖所示,$
?
?$在Rt△ACD中,$?
?$因?yàn)閟inC = \frac {AD}$?
?$所以AD= bsinC$?
?$S_{△ABC}= \frac {1}{2}a×AD=\frac {1}{2}absinC$?
?$同理可得, S_{△ABC}= \frac {1}{2}acsinB=\frac {1}{2}bcsinA$?
?$S_{△ABC}= \frac {1}{2}absinC =\frac {1}{2}acsinB=\frac {1}{2}bcsinA$?
?$(2)在Rt△ABD中,$?
?$因?yàn)閏=2 ,∠B=60°$?
?$所以AD=c×sin_{60}°=\sqrt{3},BD= c.cos_{60}°= 1$?
?$在Rt△ACD中,$?
?$因?yàn)锳D=\sqrt{3},∠C= 45°$?
?$所以CD= AD=\sqrt{3},b=\frac {AD}{sin_{45}°}=\sqrt{6}$?
?$所以a= BD+ CD=1+\sqrt{3}$?
?$S_{△ABC}=\frac {1}{2}a×AD=\frac {3+\sqrt{3}}{2}$?
?$因?yàn)镾_{△ABC}=\frac {1}{2}bcsinA$?
?$所以sinA=\frac {2S_{△ABC}}{bc}=\frac {\sqrt{2}+\sqrt{6}}{4}$?