\(\widehat{ABC}=60^0\Rightarrow\widehat{ACB}=90^0-60^0=30^0\)
Trong tam giác vuông ABH ta có:
\(sin\widehat{ABC}=\dfrac{AH}{AB}\Rightarrow AB=\dfrac{AH}{sin\widehat{ABC}}=\dfrac{\sqrt{3}}{sin60^0}=2\)
Trong tam giác vuông ACH:
\(sin\widehat{ACB}=\dfrac{AH}{AC}\Rightarrow AC=\dfrac{AH}{sin\widehat{ACB}}=\dfrac{\sqrt{3}}{sin30^0}=2\sqrt{3}\)
Pitago: \(BC=\sqrt{AB^2+AC^2}=4\)
