Hình thang ABCD (AB // CD) có góc A - góc D = 20 độ, góc B = 2 góc C. Tính các

Ta có: \[\widehat A = 100^\circ ;\widehat B = 120^\circ ;\widehat C = 60^\circ ;\widehat D = 80^\circ \](2 góc kề tổng bằng 180°)

\[ \Rightarrow \left\{ \begin{array}{l}2\widehat A = 200^\circ \\A - D = 20^\circ \end{array} \right. \Rightarrow \left\{ \begin{array}{l}\widehat A = 100^\circ \\\widehat D = \widehat A - 20^\circ = 100^\circ - 20^\circ = 80^\circ \end{array} \right.\]

Lại có: \[\left\{ \begin{array}{l}\widehat B + \widehat C = {180^^\circ }\\\widehat B = 2\widehat C\end{array} \right.\](2 góc kề có tổng bằng 180°)

\[ \Rightarrow \left\{ \begin{array}{l}\widehat B = 2\widehat C\\2\widehat C + \widehat C = 180^\circ \end{array} \right. \Rightarrow \left\{ \begin{array}{l}\widehat B = 2\widehat C\\3\widehat C = 180^\circ \end{array} \right. \Rightarrow \left\{ \begin{array}{l}\widehat B = 120^\circ \\\widehat C = 60^\circ \end{array} \right.\]

Vậy \[\widehat A = 100^\circ ;\widehat B = 120^\circ ;\widehat C = 60^\circ ;\widehat D = 80^\circ \].