Bài 3:
Ta có: \(\widehat{AOC}+\widehat{BOC}=180^0\)(hai góc kề bù)
=>\(\widehat{AOC}+50^0=180^0\)
=>\(\widehat{AOC}=130^0\)
OM là phân giác của góc AOC
=>\(\widehat{AOM}=\dfrac{\widehat{AOC}}{2}=65^0\)
Ta có: \(\widehat{AOM}+\widehat{BOM}=180^0\)(hai góc kề bù)
=>\(\widehat{BOM}=180^0-65^0=115^0\)
Bài 4:
a: Ta có: \(\widehat{AOC}+\widehat{BOC}=180^0\)(hai góc kề bù)
=>\(\widehat{BOC}+50^0=180^0\)
=>\(\widehat{BOC}=130^0\)
b: Ta có: \(\widehat{AOC}+\widehat{COD}+\widehat{DOB}=180^0\)
=>\(\widehat{COD}+50^0+40^0=180^0\)
=>\(\widehat{COD}=90^0\)
=>OC\(\perp\)OD
c: OI là phân giác của góc COD
=>\(\widehat{COI}=\dfrac{\widehat{COD}}{2}=\dfrac{90^0}{2}=45^0\)
