Do ABCD nội tiếp \(\Rightarrow A+C=180^0\Rightarrow2C+C=180^0\)
\(\Rightarrow C=60^0\)
\(\Rightarrow A=120^0\)
Vì \(\diamond ABCD\) nội tiếp nên ta có:
\(\left\{{}\begin{matrix}\widehat{A}+\widehat{C}=180^\circ\\\widehat{A}=2\widehat{C}\end{matrix}\right.\Rightarrow3\widehat{C}=180^\circ\Rightarrow\widehat{C}=60^\circ\)
\(\Rightarrow\widehat{A}=180^\circ-\widehat{C}=180^\circ-60^\circ=120^\circ\)
Vậy chọn phương án C.
Tứ giác `ABCD` nội tiếp
\(\Rightarrow\widehat{A}+\widehat{C}=180^o\\ \Rightarrow2\widehat{C}+\widehat{C}=180\\ \Rightarrow3\widehat{C}=180^o\\ \Rightarrow\widehat{C}=60^o\\ \Rightarrow\widehat{A}=2\widehat{C}=2.60=120^o\)
`=>C`
Do ABCD nội tiếp
⇒A+C=1800⇒2C+C=1800
⇒C=600
⇒A=1200
Vậy đáp án chính xác là C.1200
\(\widehat{A}+\widehat{C}=\dfrac{360^o}{2}=180^o\\ Mà:\widehat{A}=2.\widehat{C}\\ \Rightarrow2\widehat{C}+\widehat{C}=180^o\\ \Rightarrow\widehat{A}=120^o\)