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