Giải thích các bước giải:

a.Ta có: $\Delta ABC$ vuông tại $A$

$\to BC^2=AB^2+AC^2$

$\to AC^2=BC^2-AB^2$

$\to AC=\sqrt{BC^2-AB^2}=3\sqrt3$

 b.Xét $\Delta ABC,\Delta EBF$ có:

Chung $\hat B$

$\widehat{BAC}=\widehat{BEF}(=90^o)$

$\to \Delta ABC\sim\Delta EBF(g.g)$

$\to \dfrac{BA}{BE}=\dfrac{BC}{FB}$

$\to \Delta BEA\sim\Delta BFC(c.g.c)$

$\to \widehat{BEA}=\widehat{BFC}$

$\to AC=AF\cdot \dfrac{AC}{AF}$

$\to AC=AF\tan\widehat{AFC}$

$\to AC=AF\tan\widehat{BFC}$

$\to AC=AF\tan\widehat{BEA}$