Ta có : \(\widehat{A_2}=180^o-\widehat{B}-\widehat{D_2}\Rightarrow\widehat{A_2}=100^o-\widehat{B}\)
\(\widehat{A_1}=180^o-\widehat{D_1}-\widehat{C}\)
\(=80^o-\widehat{C}\)
\(\Rightarrow100^o+\widehat{B}=80^o-\widehat{C}\)
\(\Rightarrow100^o-1,5.\widehat{C}=80^o-\widehat{C}\)
\(\Rightarrow100^o-80^o=-\widehat{C}+1.5.\widehat{C}\)
\(\Rightarrow20^o=\dfrac{1}{2}.\widehat{C}\Rightarrow\widehat{C}=40^o\)
Thay \(\widehat{C}=40^o\) vào \(\widehat{A}_1\), ta có:
\(\widehat{A_1}=80^o-40^o=40^o\)
Ta có \(\widehat{A_1}=\widehat{A_2}\Rightarrow\widehat{A}=40^o.2=80^o\)
\(\Rightarrow\widehat{B}=180^o-\left(\widehat{A}+\widehat{C}\right)\)
\(=180^o-\left(80^o+40^o\right)=60^o\)
Vậy \(\left\{{}\begin{matrix}\widehat{A}=80^o\\\widehat{B}=60^o\\\widehat{C}=40^o\end{matrix}\right.\)
~ Học tốt ~
