Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\Delta ABC;\widehat {BAC} = {80^0}\\
\Rightarrow \widehat A + \widehat B + \widehat C = {180^0}\\
\Rightarrow \widehat B + \widehat C = {180^0} - \widehat A\\
\Rightarrow \widehat B + \widehat C = {100^0}\left( 1 \right)
\end{array}$
a) Ta có:
$\widehat B - \widehat C = {20^0}\left( 2 \right)$
Lấy vế với vế của $(1)$ cộng với $(2)$ ta có: $2\widehat B = {120^0} \Rightarrow \widehat B = {60^0}$
Từ $\left( 2 \right) \Rightarrow \widehat C = \widehat B - {20^0} = {40^0}$
Vậy $\widehat B = {60^0};\widehat C = {40^0}$
b) Ta có:
$\begin{array}{l}
\widehat B:11 = \widehat C:9\\
\Leftrightarrow \dfrac{{\widehat B}}{{11}} = \dfrac{{\widehat C}}{9}\\
\Rightarrow \dfrac{{\widehat B}}{{11}} = \dfrac{{\widehat C}}{9} = \dfrac{{\widehat B + \widehat C}}{{11 + 9}} = \dfrac{{{{100}^0}}}{{20}} = {5^0}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{\widehat B}}{{11}} = {5^0} \Rightarrow \widehat B = {55^0}\\
\dfrac{{\widehat C}}{9} = {5^0} \Rightarrow \widehat C = {45^0}
\end{array} \right.
\end{array}$
Vậy $\widehat B = {55^0};\widehat C = {45^0}$
c) Ta có:
$\begin{array}{l}
2\widehat B = 3\widehat C\\
\Leftrightarrow \dfrac{{\widehat B}}{3} = \dfrac{{\widehat C}}{2}\\
\Rightarrow \dfrac{{\widehat B}}{3} = \dfrac{{\widehat C}}{2} = \dfrac{{\widehat B + \widehat C}}{{3 + 2}} = \dfrac{{{{100}^0}}}{5} = {20^0}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{\widehat B}}{3} = {20^0} \Rightarrow \widehat B = {60^0}\\
\dfrac{{\widehat C}}{2} = {20^0} \Rightarrow \widehat C = {40^0}
\end{array} \right.
\end{array}$
Vậy $\widehat B = {60^0};\widehat C = {40^0}$
