a)Xét ΔABC vuông A,đường cao AH:
+)AB²=BH.BC
⇔AB²=4.(BH+HC)=4.(4+6)
⇒AB=√(4.10) =2√10 (cm)
+)AC²=CH.BC
⇔AC²=6.(BH+HC)=6.(4+6)
⇒AC=√(6.10) =2√15 (cm)
+)AH²=CH.BH
⇒AH=√(4.6) =2√6 (cm)
b)
Ta có : AM=$\frac{AC}{2}$ =√15
ΔAMB ⊥ A có tan AMB^=$\frac{AB}{AM}$ =$\frac{2V10}{V15}$
=>AMB^≈59·