Từ $A$ kẻ đường cao $AH$
Ta có:
$AH = BH.\tan30^o =\dfrac{BH\sqrt3}{3}$
$AH = CH.\tan45^o = CH = BC - BH$
$\Rightarrow \dfrac{BH\sqrt3}{3} = 16 - BH$
$\Rightarrow BH = \dfrac{16}{1 + \dfrac{\sqrt3}{3}}= 24- 8\sqrt3 \, cm$
$\Rightarrow CH = 8\sqrt3 - 8 \, cm$
Ta được:
$AB = \dfrac{BH}{\cos30^o}=\dfrac{24 - 8\sqrt3}{\dfrac{\sqrt3}{2}}=16\sqrt3 - 16 \, cm$
$AC = \dfrac{CH}{\cos45^o}=\dfrac{8\sqrt3 - 8}{\dfrac{\sqrt2}{2}}= 16\sqrt{2 - \sqrt3}\, cm$
