Question

[Nguyễn Trần Như Hằng]

giúp mình vs mai nộp rồi

Answer 1

[Nguyễn Huy Tú]

Giải:

Ta có: \(\widehat{A_1}+\widehat{A_2}=180^o\) ( kề bù )

Mà \(\widehat{A_1}-\widehat{A_2}=60^o\)

\(\Rightarrow\widehat{A_1}=\left(180^o+60^o\right):2=120^o\)

\(\Rightarrow\widehat{A_2}=180^o-\widehat{A_1}=180^o-120^o=60^o\)

Vì a // b nên \(\widehat{B_1}=\widehat{A_1}=120^o\) ( so le trong )

\(\widehat{B_2}=\widehat{A_2}=60^o\) ( so le trong )

Vậy \(\widehat{B_1}=120^o,\widehat{B_2}=60^o\)

Answer 2

[Aki Tsuki]

GT: a // b ; \(\widehat{A_1}\) - \(\widehat{A_2}\) = 60^o

KL : \(\widehat{B_1}\) = ? ; \(\widehat{B_2}\) = ?

Ta có: \(\widehat{A_1}\) - \(\widehat{A_2}\) = 60^o (gt) (1)

và \(\widehat{A_1}\) + \(\widehat{A_2}\) = 180^o ( 2 góc kề bù) (2)

Từ (1) và (2)

\(\Rightarrow\) \(\widehat{A_1}\) = \(\frac{180^o+60^o}{2}\) = 120^o

^{\(\widehat{A_2}\)} = \(\frac{180^o-60^o}{2}\) = 60^o

Vì a // b (gt) nên:

\(\Rightarrow\) \(\widehat{A_1}\) = \(\widehat{B_1}\) = 120^o ( cặp góc so le trong)

\(\widehat{A_2}\) = \(\widehat{B_2}\) = 60^o ( cặp góc so le trong)

Vậy \(\widehat{B_1}\) = 120^o ; \(\widehat{B_2}\) = 60^o