a: Xet ΔABC có
\(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{8^2+5^2-BC^2}{2\cdot8\cdot5}=\dfrac{1}{2}\)
=>\(89-BC^2=40\)
=>BC=7cm
b: \(S_{ABC}=\dfrac{1}{2}\cdot8\cdot5\cdot sin60=10\sqrt{3}\left(cm^2\right)\)
c: \(S_{ABC}=\dfrac{a\cdot b\cdot c}{4RR}\)
=>\(10\sqrt{3}=\dfrac{8\cdot5\cdot7}{4R}\)
=>70/R=10căn 3
=>R=7/3căn 3(cm)