Question

Answer 1

a, Theo định lí cos:

\(AC^2=BC^2+AB^2-2AB.BC.cosB=57\)

\(\Rightarrow AC=\sqrt{57}\left(cm\right)\)

Answer 2

b, \(S_{ABC}=\dfrac{1}{2}.AB.BC.sinB=\dfrac{1}{2}.7.8.sin60^o=14\sqrt{3}\left(cm^2\right)\)

Mặt khác lại có \(S_{ABC}=\dfrac{1}{2}p.r\Rightarrow r=\dfrac{2.S_{ABC}}{p}=\dfrac{2.14\sqrt{3}}{\dfrac{7+8+\sqrt{57}}{2}}=\dfrac{56\sqrt{3}}{15+\sqrt{57}}\left(cm\right)\)