$解：过点C作CD⊥AB于点D$
$∵∠A=30°，AC=\sqrt {6}，cosA=cos{30}°=\frac {AD}{AC}=\frac {\sqrt {3}}2$
$∴AD=\frac {3\sqrt {2}}2$
$∵sinA=sin{30}°=\frac {CD}{AC}=\frac 1 2$
$∴CD=\frac {\sqrt {6}}2$
$∵tanB=tan{45}°=\frac {CD}{BD}=1$
$∴BD=\frac {\sqrt {6}}2$
$∴AB=AD+BD=\frac {3\sqrt {2}+\sqrt {6}}2$
$在Rt△BCD中，由勾股定理，得$
$BC=\sqrt {{CD}^2+{BD}^2}=\sqrt {3}$
