$\Delta$ AHC vuông tại H có:
$AH=HC.\tan C$
$\Rightarrow HC=\dfrac{6}{\frac{2}{3}}=9(cm)$
$\Rightarrow AC=\sqrt{AH^2+HC^2}=3\sqrt{13}(cm)$
$\Delta$ ABC vuông tại A, $AH\bot BC$ có:
$AC^2=HC.BC$
$\Leftrightarrow BC=13(cm)$
$\Rightarrow AB=\sqrt{BC^2-AC^2}=2\sqrt{13}$
$\widehat{A}=90^o$
$\tan C=\dfrac{2}{3}\Rightarrow \widehat{C}\approx 33^o41'$
$\Rightarrow \widehat{B}=90^o-33^o44'=56^o16'$