In the inner region of the right angle aob, a ray oc is drawn, find the angle between the bisectors of the angles AOC and BOC.

Let the value of the angle AOC = X0, then the value of the angle BOC will be equal to (90 – X) 0.

Since OD is the bisector of the angle AOC, then the angle AOC = DOA = AOC / 2 = (X / 2) 0.

OE is the bisector of the angle BOC, then the angle COE = BOE = COE / 2 = (90 – X) / 2 = (45 – X / 2) 0.

The angle between the bisectors OD and OE will be equal to:

DOE = DOC + COE = (X / 2) + (45 – X / 2) = 45.

Answer: The angle between the bisectors of the angles is 45.