Question

line ob bisects angle nod and line oa and line oc trisects angle nod and angle aob is 20 degrees less than angle noa. find the measure of angle nod and boc

Answer 1

Answer:

$m\angle NOD=120^{\circ}\\ \\m\angle BOC=20^{\circ}$

Step-by-step explanation:

If lines OA and OC trisect angle NOD, then

$m\angle NOA=m\angle AOC=m\angle COD=(2x)^{\circ}$

By angle addition postulate:

$m\angle NOD=m\angle NOA+m\angle AOC+m\angle COD=(2x)^{\circ}+(2x)^{\circ}+(2x)^{\circ=6x^{\circ}}$

If line OB bisects ange NOD, then

$m\angle NOB=m\angle BOD=\dfrac{(6x)^{\circ}}{2}=(3x)^{\circ}$

Angle AOB is 20 degrees less than angle NOA, then

$m\angle NOA-m\angle AOB=20^{\circ}$

By angle addition postulate,

$m\angle NOB=m\angle NOA+m\angle AOB\Rightarrow (2x)^{\circ}+m\angle AOB=(3x)^{\circ}\\ \\m\angle AOB=x^{\circ}$

Thus,

$(2x)^{\circ}-x^{\circ}=20^{\circ}\\ \\x=20^{\circ}$

and

$m\angle NOD=6\cdot 20^{\circ}=120^{\circ}\\ \\m\angle BOC=x^{\circ}=20^{\circ}$