Đáp án:
\[\left\{ \begin{array}{l}
a = 10\sqrt {19} \left( {cm} \right)\\
\widehat B = 36,6^\circ \\
\widehat C = 83,4^\circ
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{a^2} = {b^2} + {c^2} - 2bc.\cos A = {30^2} + {50^2} - 2.30.50.\cos 60^\circ = 1900\\
\Rightarrow a = 10\sqrt {19} \,\,\,\left( {cm} \right)\\
\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\\
\Leftrightarrow \dfrac{{10\sqrt {19} }}{{\sin 60^\circ }} = \dfrac{{30}}{{\sin B}} = \dfrac{{50}}{{\sin C}}\\
\Leftrightarrow \left\{ \begin{array}{l}
\sin B = \dfrac{{30.\sin 60^\circ }}{{10\sqrt {19} }}\\
\sin C = \dfrac{{50.\sin 60^\circ }}{{10\sqrt {19} }}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\widehat B = 36,6^\circ \\
\widehat C = 83,4^\circ
\end{array} \right.
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
a = 10\sqrt {19} \left( {cm} \right)\\
\widehat B = 36,6^\circ \\
\widehat C = 83,4^\circ
\end{array} \right.\)
