Diện tích tam giác ABC là:
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA\)
\(=\dfrac{1}{2}\cdot7\cdot5\cdot\dfrac{\sqrt{3}}{2}\)
\(=\dfrac{35\sqrt{3}}{4}\)
Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{7^2+5^2-BC^2}{2\cdot7\cdot5}=cos60=\dfrac{1}{2}\)
=>\(49+25-BC^2=7\cdot5=35\)
=>\(BC^2=49+25-35=39\)
=>\(BC=\sqrt{39}\)
Độ dài đường cao kẻ từ A xuống BC là:
\(h_A=\dfrac{2\cdot S}{BC}=\dfrac{2\cdot\dfrac{35\sqrt{3}}{4}}{\sqrt{39}}=\dfrac{35\sqrt{3}}{2}:\sqrt{39}=\dfrac{35\sqrt{3}}{2\sqrt{39}}=\dfrac{35}{2\sqrt{13}}=\dfrac{35\sqrt{13}}{26}\)
Xét ΔABC có \(\dfrac{BC}{sinA}=2R\)
=>\(2R=\sqrt{39}:\dfrac{\sqrt{3}}{2}=\sqrt{39}\cdot\dfrac{2}{\sqrt{3}}=2\sqrt{13}\)
=>\(R=\sqrt{13}\)
Nửa chu vi tam giác ABC là:
\(p=\dfrac{AB+AC+BC}{2}=\dfrac{7+5+\sqrt{39}}{2}=\dfrac{12+\sqrt{39}}{2}\)
\(S=p\cdot r\)
=>\(r=\dfrac{35\sqrt{3}}{4}:\dfrac{12+\sqrt{39}}{2}=\dfrac{35\sqrt{3}}{2\left(12+\sqrt{39}\right)}\)