a, Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
\(\Leftrightarrow\widehat{A}=180^0-\widehat{B}-\widehat{C}=180^0-22^0-90^0=68^0\)
b, \(\widehat{MNP}=180^0-85^0=95^0\)
Ta có: \(\widehat{M}+\widehat{P}+\widehat{MNP}=180^0\)
\(\Leftrightarrow\widehat{P}=180^0-\widehat{M}-\widehat{MNP}=180^0-25^0-95^0=60^0\)
c, Vì Bx // AC \(\Rightarrow\widehat{C}=55^0\) (2 góc so le trong)
Ta có: \(\widehat{A}+\widehat{ABC}+\widehat{C}=180^0\)
\(\Leftrightarrow\widehat{A}=180^0-\widehat{ABC}-\widehat{C}=180^0-83^0-55^0=42^0\)
Vì Bx // AC \(\Rightarrow\widehat{ABx}=\widehat{A}=42^0\) (2 góc so le trong)
d, Ta có: \(x+3x+x+10=180^0\)
\(\Leftrightarrow5x+10=180\)
\(\Leftrightarrow5x=170\)
\(\Leftrightarrow x=34\)
Vậy \(\widehat{E}=34^0,\widehat{D}=3.34=102^0,\widehat{F}=34+10=44^0\)
a: x=90-22=68 độ
b:y=180-25-(180-85)=-25+85=60 độ
c: x=180-83-55=42 độ
=>y=42 độ
=>z=180-42-55=83 độ
