`@` `\text {Ans}`
`\downarrow`
`20,`
Ta có:
\(\widehat{EDy}+\widehat{EDC}=180^0\left(\text{kề bù}\right)\)
`=>`\(130^0+\widehat{EDC}=180^0\)
`=>`\(\widehat{EDC}=180^0-130^0=50^0\)
Ta có:
\(\widehat{DEB}+\widehat{EDC}=180^0\left(\text{2 góc trong cùng phía bù nhau}\right)\)
`=>`\(\widehat{DEB}+50^0=180^0\)
`=>`\(\widehat{DEB}=180^0-50^0=130^0\)
Vậy, số đo $\widehat {DEB}$ là `130^0.`