Answer :

The figure is attached below:

BN and AM are the angle bisectors of angle B and A respectively.

Given ∠C = 70°

In a triangle we know sum of interior angles is 180°

In ΔABC

∠A + ∠B + ∠C = 180°

∠A + ∠B = 180° - 70°

∠A + ∠B = 110°

Now in ΔAOB

AO is the bisector of ∠A

BO is the bisector of ∠B

∴ ∠OAB = ∠A/2 and ∠OBA = ∠B/2

∠OAB + ∠OBA + ∠AOB = 180°

∠A/2 + ∠B/2 + ∠AOB = 180°

⇒ ∠AOB = 180° - (∠A + ∠B)/2

⇒ ∠AOB = 180° - 110°/2 = 180° - 55°

⇒ ∠AOB = 125°

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