here is a triangle (ABC),A^ =80 and O is a point in triangle (not clear) ,?

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NOTE

I copied the problem incorrectly.

I have AB = BC

instead of AB = AC.

I'm tired.

Off hand, I don't see a way to do this if AB = AC.

Good luck.

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Given:

m∠A = 80°

Point O is in the interior of ΔABC

m∠CBO = 10°

m∠OCB = 30°

AB = BC

Find:

m∠AOB

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1. m∠BOC + m∠OBC + m∠OCB = 180° so that m∠BOC = 140°

2. ΔABC is isosceles by definition, since AB = BC

3. m∠BCA = m∠BAC = 80°, since base angles of isosceles Δ's are congruent

4. m∠BAC + m∠BCA + m∠ABC = 180°

5. m∠ABC = 180° − m∠BAC − m∠BCA = 20°

6. m∠ABC = m∠ABO + m∠OBC

                   20° = m∠ABO + 10°

       m∠ABO = 10°

7. OB = OB

8. ΔBOA ≅ ΔBOC by SAS

9. ∠AOB ≅ ∠COB by corresponding parts of congruent Δ's

10. Therefore m∠AOB = 140°

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