Question

help me

 

Answer 1

Bài nào vậy bạn

Answer 2

bài 3

 

Answer 3

1.

Ta có \(\widehat{xOy}=\widehat{x'Oy'}=65^0\left(đối.đỉnh\right)\)

Ta có \(\widehat{xOy}+\widehat{xOy'}=180^0\left(kề.bù\right)\Rightarrow\widehat{xOy'}=180^0-65^0=115^0\)

Ta có \(\widehat{xOy'}=\widehat{x'Oy}=115^0\left(đối.đỉnh\right)\)

2.

a, Ta có \(\widehat{A_2}=\widehat{A_1}=52^0\left(đối.đỉnh\right)\)

Vì a//b nên \(\widehat{A_1}=\widehat{B_2}=52^0\left(so.le.trong\right)\)

Ta có \(\widehat{B_2}+\widehat{B_1}=180^0\left(kề.bù\right)\Rightarrow\widehat{B_1}=180^0-52^0=128^0\)

Ta có \(\widehat{B_1}=\widehat{B_3}=128^0\left(đối.đỉnh\right)\)

3.

Kẻ Oa//m//n

\(\Rightarrow\widehat{mMO}=\widehat{aOM}=35^0;\widehat{nNO}=\widehat{aON}=45^0\left(so.le.trong\right)\)

\(\Rightarrow\widehat{MON}=\widehat{mMO}+\widehat{nNO}=45^0+35^0=80^0\)