Question

Cho đường thẳng c cắt hai đường thẳng a và b tại A và B , biết \(\widehat{A}_1+\widehat{A}_2+\widehat{A_3}=310^0;\widehat{B_2}-\widehat{B_1}=80^0.\)Tính các góc \(\widehat{A_1}\) ; \(\widehat{A_2}\) ; \(\widehat{A_3}\) ; \(\widehat{A_4}\) ; \(\widehat{B_1}\); \(\widehat{B_2}\) ; \(\widehat{B_3}\) ; \(\widehat{B_4}\)

Answer 1

- Theo đề bài ta có:

\(\widehat{A1}+\widehat{A2}+\widehat{A3}=310^o\)

=> \(\widehat{A4}=360^o-310^o=50^o\)

- Ta có:

+ \(\widehat{A3}\) kề bù \(\widehat{A4}\)

=> \(\widehat{A3}\) = 180^o - 50^o = 130^o

^{+ \(\widehat{A4}\)} Đối đỉnh \(\widehat{A2}\)

=> \(\widehat{A2}\) = 50^o.

+ \(\widehat{A3}\) đối đỉnh \(\widehat{A1}\)

=> \(\widehat{A1}\) = 130^o.

- Lại có:

\(\widehat{B2}-\widehat{B1}=80^o\) và \(\widehat{B2}+\widehat{B1}=180^o\)( 2 góc kề bù)

=> \(\widehat{B2}=\dfrac{\left(180^o-80^o\right)}{2}=50^o\)

=> \(\widehat{B1}=180^o-50^o=130^o\)

=> \(\widehat{B3}=130^o\)( B3 đối đỉnh B1)

=> \(\widehat{B4}=50^o\)( B4 đối đỉnh B2).

Answer 2

Ta có: \(\widehat{A}1+\widehat{A2}+\widehat{A3}+\widehat{A4}=360^0\)

Mà\(\widehat{A1}+\widehat{A2}+\widehat{A3}=310^0\)

\(\Rightarrow\widehat{A4}=50^0,\widehat{A2}=50^0\)(2 góc đối đỉnh)

\(\Rightarrow\widehat{A3}=\widehat{A1}=130^0\)(Kề bù vời\(\widehat{A2},\widehat{A4}\))

\(\widehat{B2}-\widehat{B1}=80^0\)Mà\(\widehat{B1}+\widehat{B2}=180^0\)\(\Rightarrow\widehat{B2}=\widehat{B4}=130^0\)(2 góc đối đỉnh)

\(\Rightarrow\widehat{B1}=\widehat{B3}=5o^0\)(Kề bù vời\(\widehat{B1},\widehat{B3}\))

Vậy\(\widehat{A2=\widehat{A4}=\widehat{B1}}=\widehat{B3}=30^0\)

\(\widehat{B2}=\widehat{B4}=\widehat{A1=\widehat{A3}=}130^0\)