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Diện tích tam giác ABC là:
\(S_{ABC}=\dfrac{1}{2}\cdot BA\cdot BC\cdot sinABC\)
\(=\dfrac{1}{2}\cdot5\cdot7\cdot sin120=\dfrac{35\sqrt{3}}{4}\)
Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(\dfrac{5^2+7^2-AC^2}{2\cdot5\cdot7}=cos120=\dfrac{-1}{2}\)
=>\(25+49-AC^2=-35\)
=>\(AC^2=25+49+35=109\)
=>\(AC=\sqrt{109}\)
Kẻ AH\(\perp\)BC
=>\(h_A=AH\)
\(S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC\)
=>\(\dfrac{1}{2}\cdot AH\cdot7=\dfrac{35\sqrt{3}}{4}\)
=>\(AH\cdot3,5=\dfrac{35\sqrt{3}}{4}\)
=>\(AH=\dfrac{10\sqrt{3}}{4}=\dfrac{5}{2}\sqrt{3}\)
Xét ΔABC có \(\dfrac{AC}{sinB}=2R\)
=>\(2R=\dfrac{\sqrt{109}}{sin120}=\sqrt{109}\cdot\dfrac{2}{\sqrt{3}}\)
=>\(R=\sqrt{\dfrac{109}{3}}=\dfrac{\sqrt{327}}{3}\)
Xét ΔABC có \(cosA=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
\(=\frac{8^2+10^2-7^2}{2\cdot8\cdot10}=\frac{115}{16\cdot10}=\frac{115}{5\cdot32}=\frac{23}{32}\)
=>\(\sin A=\sqrt{1-\left(\frac{23}{32}\right)^2}=\frac{3\sqrt{55}}{32}\)
\(S_{ABC}=\frac12\cdot b\cdot c\cdot\sin A\)
\(=\frac12\cdot8\cdot10\cdot\frac{3\sqrt{55}}{32}=\frac{80}{64}\cdot3\sqrt{55}=\frac54\cdot3\sqrt{55}=\frac{15\sqrt{55}}{4}\)
Xét ΔABC có \(\frac{BC}{\sin A}=2R\)
=>\(2R=7:\frac{3\sqrt{55}}{32}=7\cdot\frac{32}{3\sqrt{55}}=\frac{224}{3\sqrt{55}}\)
=>\(R=\frac{112}{3\sqrt{55}}\)
\(p=\frac{a+b+c}{2}=\frac{7+8+10}{2}=\frac{25}{2}=12,5\)
S=p*r
=>\(r=\frac{15\sqrt{55}}{4}:12,5=\frac{15\sqrt{55}}{4\cdot12,5}=\frac{15\sqrt{55}}{50}=\frac{3\sqrt{55}}{10}\)
\(S=\frac12\cdot a\cdot h_{a}\)
=>\(\frac12\cdot7\cdot h_{a}=\frac{15\sqrt{55}}{4}\)
=>\(h_{a}=\frac{15\sqrt{55}}{4}:\frac72=\frac{15\sqrt{55}}{4}\cdot\frac27=\frac{15\sqrt{55}}{14}\)
Xét ΔABC có \(m_{a}\) là độ dài đường trung tuyến
nên \(m_{a}^2=\frac{b^2+c^2}{2}-\frac{a^2}{4}=\frac{8^2+10^2}{2}-\frac{7^2}{4}=69,75\)
=>\(m_{a}=\sqrt{69,75}=\sqrt{\frac{279}{4}}=\frac{3\sqrt{31}}{2}\)