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Trả lời:
$a,$ Áp dụng định lí cosin:
$BC^2=AB^2+AC^2-2.AB.AC.cosA$
$BC^2=4^2+8^2-2.4.8.cos60^o=48$
$BC=4\sqrt{3}\,(cm)$
$b,$
$S_{ΔABC}=\dfrac{1}{2}.AB.AC.sinA\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}.4.8.sin60^o=8\sqrt{3}\,(cm^2)$
Mặt khác: $S_{ΔABC}=\dfrac{AB.AC.BC}{4R}$
$⇒4R=\dfrac{4.8.4\sqrt{3}}{8\sqrt{3}}=16$
$⇒R=4\,(cm)$.