a; Xét ΔABC vuông tại A có \(\tan B=\dfrac{AC}{AB}=\dfrac{4}{3}\)
nên \(\widehat{B}\simeq53^0\)
=>\(\widehat{C}=37^0\)
b: \(\dfrac{HB}{HC}=\left(\dfrac{AB}{AC}\right)^2=\dfrac{9}{16}\)
c: \(\widehat{BDA}+\widehat{HAD}=90^0\)
\(\widehat{BAD}+\widehat{CAD}=90^0\)
mà \(\widehat{HAD}=\widehat{CAD}\)
nên \(\widehat{BDA}=\widehat{BAD}\)
hay ΔABD cân tại B