Answer:
[tex]m\angle NOD=120^{\circ}\\ \\m\angle BOC=20^{\circ}[/tex]
Step-by-step explanation:
If lines OA and OC trisect angle NOD, then
[tex]m\angle NOA=m\angle AOC=m\angle COD=(2x)^{\circ}[/tex]
By angle addition postulate:
[tex]m\angle NOD=m\angle NOA+m\angle AOC+m\angle COD=(2x)^{\circ}+(2x)^{\circ}+(2x)^{\circ=6x^{\circ}}[/tex]
If line OB bisects ange NOD, then
[tex]m\angle NOB=m\angle BOD=\dfrac{(6x)^{\circ}}{2}=(3x)^{\circ}[/tex]
Angle AOB is 20 degrees less than angle NOA, then
[tex]m\angle NOA-m\angle AOB=20^{\circ}[/tex]
By angle addition postulate,
[tex]m\angle NOB=m\angle NOA+m\angle AOB\Rightarrow (2x)^{\circ}+m\angle AOB=(3x)^{\circ}\\ \\m\angle AOB=x^{\circ}[/tex]
Thus,
[tex](2x)^{\circ}-x^{\circ}=20^{\circ}\\ \\x=20^{\circ}[/tex]
and
[tex]m\angle NOD=6\cdot 20^{\circ}=120^{\circ}\\ \\m\angle BOC=x^{\circ}=20^{\circ}[/tex]
