a, Ta có: \(a^2=b^2+c^2-2bc.cosA\)
\(\Leftrightarrow a^2=20^2+35^2+2.20.35.cos50^0\)
\(\Leftrightarrow a\approx27\)
Có: \(m^2_b=\frac{2\left(a^2+c^2\right)-b^2}{4}=\frac{2.\left(27^2+35^2\right)-20^2}{4}\)
\(\Leftrightarrow m_b^2=877\Leftrightarrow m_b=\sqrt{877}\approx29,6\)
b, Ta có: \(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{27^2+35^2-20^2}{2.27.35}=\frac{37}{45}\)
\(\Leftrightarrow\widehat{B}\approx34,7^0\)
Có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{C}=180^0-50^0-34,7^0=95,3^0\)
Ta có: \(\frac{a}{sinA}=2R\Leftrightarrow R=\frac{a}{2.sinA}=\frac{27}{2.sin50^0}\approx17,6\)
Good luck!
