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Trả lời:
Kẻ đường cao $AH\perp BC$
$\widehat{ABH}=\widehat{ACH}=30^o$
$⇒\sin ABH=\dfrac{AH}{AB}=\sin 30^o=\dfrac{1}{2}$
$⇒AH=AB.\dfrac{1}{2}=\dfrac{5}{2}\,(cm)$
Áp dụng định lí Py-ta-go:
$BH=\sqrt{AB^2-AH^2}=\sqrt{5^2-\dfrac{25}{4}}=\dfrac{5\sqrt{3}}{2}$
$⇒BC=2.BH=5\sqrt{3}\,(cm)$
$⇒S_{ΔABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.\dfrac{5}{2}.5\sqrt{3}=\dfrac{25\sqrt{3}}{4}\,(cm^2)$.