Giải thích các bước giải:
a.Ta có $AH\perp BC\to AH=\sqrt{AB^2-BH^2}=4$
b.Ta có $AH\perp BC\to HC=\sqrt{AC^2-AH^2}=\dfrac{16}{3}$
Xét $\Delta AHB,\Delta AHC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\dfrac{HA}{HC}=\dfrac{HB}{HA}(=\dfrac34)$
$\to \Delta HAB\sim\Delta HCA(c.g.c)$
$\to \widehat{HAB}=\widehat{HCA}$
$\to \widehat{BAC}=\widehat{BAH}+\widehat{HAC}=\widehat{HCA}+\widehat{HAC}=90^o$