Question

mọi người làm giúp mình với , mình cần gấp

Answer 1

a) Ta có: \(\widehat{AOM}=90^0\)

\(\Rightarrow MO\perp AB\Rightarrow\widehat{MOB}=90^0\)

\(\Rightarrow\widehat{NOM}=\widehat{MOB}-\widehat{BON}=90^0-35^0=55^0\)

b) Ta có: \(\widehat{AOM}=90^0,\widehat{MON}=55^0,\widehat{NOB}=35^0\)

\(\Rightarrow\widehat{AOM}>\widehat{MON}>\widehat{NOB}\)

c) Cặp góc phụ nhau: \(\widehat{BON}\) và \(\widehat{MON}\)

Các cặp góc bù nhau: \(\widehat{AOM}\) và \(\widehat{MOB}\), \(\widehat{BON}\) và \(\widehat{AON}\)

Cặp góc bằng nhau: \(\widehat{AOM}\) và \(\widehat{BOM}\)

Answer 2

\(a,\widehat{NOM}=\widehat{BOM}-\widehat{NOB}=90^0-35^0=55^0\\ b,90^0>55^0>35^0\Rightarrow\widehat{AOM}>\widehat{MON}>\widehat{NOB}\\ c,\)

Cặp góc phụ nhau: \(\widehat{MON}.và.\widehat{NOB}\)

Cặp góc bù nhau: \(\widehat{AOM}.và.\widehat{MOB};\widehat{AON}.và.\widehat{NOB}\)

Cặp góc bằng nhau: \(\widehat{AOM}=\widehat{BOM}\left(=90^0\right)\)