a) Ta có :
AOM + BON = 180độ
hay AON + MON + BOM + MON = 180
AON + BOM + 2MON = 180
mà AON + MON + BOM = AOB = 100độ
=> MON + 100 = 180
=> MON = 80độ
Tự vẽ hình nhau Đ
a) Ta có: \(\widehat{AOM}+\widehat{NOB}=90^0+90^0=180^0\)
Mà \(\widehat{AOM}=\widehat{AON}+\widehat{MON}\)
\(\widehat{NOB}=\widehat{MON}+\widehat{MOB}\)
\(\Rightarrow\widehat{AON}+\widehat{MON}+\widehat{MON}+\widehat{MOB}=180^0\)
Mà \(\widehat{AON}+\widehat{MON}+\widehat{MOB}=100^0=\widehat{AOB}\)
\(\Rightarrow\widehat{AON}+2\widehat{MON}+\widehat{MOB}=100^0+\widehat{MON}=180^0\)
\(\Rightarrow\widehat{MON}=180^0-100^0=80^0\)
b) \(\widehat{AON}=90^0-\widehat{MON}=90^0-80^0=10^0\)
Mà Oa là tia phân giác của góc AON \(\Rightarrow\widehat{AOa}=\widehat{NOa}=\frac{\widehat{AON}}{2}=\frac{10^0}{2}=5^0\)
\(\widehat{MOB}=90^0-\widehat{MON}=90^0-80^0=10^0\)
Mà Ob là tia phân giác của góc BOM \(\Rightarrow\widehat{MOb}=\widehat{Bob}=\frac{\widehat{BOM}}{2}=\frac{10^0}{2}=5^0\)
\(\Rightarrow\widehat{aOb}=\widehat{NOa}+\widehat{MON}+\widehat{MOb}=80^0+5^0+5^0=90^0\)
Vậy \(Oa\perp Ob\)