a, BC=BH+HC=8BC=BH+HC=8
Áp dụng HTL:
⎧⎪⎨⎪⎩AB2=BH⋅BC=16AC2=CH⋅BC=48AH2=CH⋅BC=12⇒⎧⎪ ⎪⎨⎪ ⎪⎩AB=4(cm)AC=4√3(cm)AH=2√3(cm){AB2=BH⋅BC=16AC2=CH⋅BC=48AH2=CH⋅BC=12⇒{AB=4(cm)AC=43(cm)AH=23(cm)
b,b, Vì K là trung điểm AC nên AK=12AC=2√3(cm)AK=12AC=23(cm)
Ta có tanˆAKB=ABAK=42√3=2√33≈tan490tanAKB^=ABAK=423=233≈tan490
⇒ˆAKB≈490