Xét tứ giác `ABCD` có :
`\hat{A} + \hat{B} + \hat{C} + \hat{D} = 360^o`
Mà `\hat{A} = 65^o` (gt)
` \hat{B} = 117^o` (gt)
`\hat{C} = 71^o` (gt)
`=> \hat{D} = 107^o`
Mà góc ngoài tại đỉnh `D` kề bù với `\hat{D}`
`=>` Tổng số đo của góc ngoài tại đỉnh `D` và `\hat{D}` là `180^o`
`=>` Số đo góc ngoài tại đỉnh `D` là `180^o - 107^o = 73^o`
Đáp án:
Giải thích các bước giải:
Xét tứ giác $ABCD$ ta có:$\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0$
$⇒\widehat{D}=360^0-(\widehat{A}+\widehat{B}+\widehat{C})=360^0-(65^0+117^0+71^0)=360^0-253^0=107^0$
$⇒$Góc ngoài tại đỉnh $D=180^0-107^0=73^0$
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