Giải:
Ta có:
\(\widehat{A}+\widehat{ACD}+\widehat{ADC}=180^0\) (Tổng ba góc của tam giác)
Hay \(50^0+\widehat{ACD}+\widehat{ADC}=180^0\)
\(\widehat{ACD}+\widehat{ADC}=180^0-50^0=130^0\)
Lại có:
\(\left\{{}\begin{matrix}\widehat{BCD}=\widehat{ACD}-\widehat{ACB}\\\widehat{BDC}=\widehat{ADC}-\widehat{ADB}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{BCD}=\widehat{ACD}-30^0\\\widehat{BDC}=\widehat{ADC}-20^0\end{matrix}\right.\)
Mặt khác:
\(\widehat{BCD}+\widehat{BDC}+\widehat{CBD}=180^0\) (Tổng ba góc của tam giác)
\(\Leftrightarrow\left(\widehat{ACD}-30^0\right)+\left(\widehat{ADC}-20^0\right)+\widehat{CBD}=180^0\)
\(\Leftrightarrow\widehat{ACD}-30^0+\widehat{ADC}-20^0+\widehat{CBD}=180^0\)
\(\Leftrightarrow\widehat{ACD}+\widehat{ADC}-30^0-20^0+\widehat{CBD}=180^0\)
\(\Leftrightarrow130^0-50^0+\widehat{CBD}=180^0\)
\(\Leftrightarrow80^0+\widehat{CBD}=180^0\)
\(\Leftrightarrow\widehat{CBD}=180^0-80^0=100^0\)
Vậy \(\widehat{CBD}=100^0\).
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