如图,在△ABC中,∠B=45°,∠C=60°,BC=6,求边AB,AC的长（结果保留根号）

∠A=180°-∠B-∠C=180°-45°-60°=75°
sin75°=sin(45°+30°)=sin45°cos30°+cos45°sin30°
=√2/2*1/2+√2/2*√3/2=(√2+√6)/4
AB/BC=sin∠C/sin∠A
AB=BC*sin∠C/sin∠A=6*sin60°/sin75°
=6*√3/2/[(√2+√6)/4]
=9√2-3√6
AC/BC=sin∠B/sin∠A
AC=BC*sin∠B/sin∠A=6*sin45°/sin75°
=6*√2/2/[(√2+√6)/4]
=6√3-6