a, $ΔABC$ vuông tại $A$
$⇒cosC=\dfrac{AC}{BC}$
$⇒BC=\dfrac{AC}{cosC}=\dfrac{14}{cos64^o}=31,9(m)$
$AB^2+AC^2=BC^2$ (Ptg)
$⇒AB=\sqrt{BC^2-AC^2}=\sqrt{31,9^2-14^2}=28,7(m)$
$\widehat{B}+\widehat{C}=90^o$ (phụ nhau)
$⇒\widehat{B}=90^o-\widehat{C}=90^o-64^o=26^o$
b, $CE$ là phân giác $\widehat{C}$
$⇒\dfrac{BE}{EA}=\dfrac{BC}{AC}=\dfrac{31,9}{14}=\dfrac{319}{140}$
$⇒\dfrac{BE}{319}=\dfrac{EA}{140}=\dfrac{BE+EA}{319+140}=\dfrac{28,7}{459}=\dfrac{287}{4590}$
$\dfrac{BE}{319}=\dfrac{287}{4590}⇒BE=\dfrac{287}{4590}.319=19,9(m)$