Sửa đề; BA=9cm
Xét ΔBAC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(9^2+8^2-AC^2=2\cdot9\cdot8\cdot cos70\)
\(\Leftrightarrow AC\simeq9.79\left(cm\right)\)
\(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)
=>\(\widehat{C}\simeq60^0\)
=>góc A=180-70-60=50 độ
b: \(S_{ABC}=\dfrac{1}{2}\cdot9\cdot8\cdot sin70\simeq33.83\left(cm^2\right)\)
p=(9+8+9,79)/2=13,395(cm)
S=p*r
=>r=2,53(cm)
AC/sinB=2*R
=>2*R=9,79/sin70
=>R=5,21(cm)
c: \(AM=\sqrt{\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}}=8.51\left(cm\right)\)
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