$解：如图，过点A作AH⊥BC于点H$
$则∠AHB=∠AHC= 90°$
$在Rt△AHC中,∠C=30°,∠AHC= 90° ,AC= 2\ \mathrm {cm}$
$∴ AH=1\ \mathrm {cm}$
$由勾股定理,得AH^2+ HC^2=AC^2$
$∴ HC=\sqrt{3}\ \mathrm {cm}(负值舍去)$
$在△ABC中,∵∠BAC= 105°,∠C= 30°$
$∴∠B= 180°-∠BAC-∠C = 45°$
$在Rt△ABH中，∵∠AHB=90°,∠B=45°,AH=1\ \mathrm {cm}$
$∴BH=1\ \mathrm {cm}$
$∴ BC=HC+ BH=(\sqrt{3}+1)\ \mathrm {cm}$
