# In triangle OAB, angle O is 30 degrees. Angle A is 85 degrees. AK-, bisector of angle A.

In triangle OAB, angle O is 30 degrees. Angle A is 85 degrees. AK-, bisector of angle A. Find the angles OKA, AKB and OBA.

Given: △ OAB, ∠O = 30 °, ∠A = 85 °, AK – bisector of angle A.
Find: ∠OKA, ∠AKB, ∠OBA.
For the theorem on the sum of the angles of a triangle, in △ OAB:
∠AOB + ∠OAB + ∠OBA = 180 °.
Hence:
∠OBA = 180 ° – (∠AOB + ∠OAB) = 180 ° – (30 ° + 85 °) = 65 °.
A bisector is a ray that divides a vertex into 2 equal angles. We have:
∠OAK = ∠OAB / 2 = 85 ° / 2 = 42.5 °.
∠BAK = ∠OAK = 42.5 °.
Behind the triangle sum theorem, in △ OAK:
∠OAK + ∠AOK + ∠OKA = 180 °.
Hence:
∠OKA = 180 ° – (∠OAK + ∠AOK) = 180 ° – (42.5 ° + 30 °) = 107.5 °.
Behind the triangle sum theorem, in △ BAK:
∠ABK + ∠BAK + ∠AKB = 180 °.
∠ABK = ∠OBA = 65 °.
Hence:
∠AKB = 180 ° – (∠ABK + ∠BAK) = 180 ° – (65 ° + 42.5 °) = 72.5 °.
Answer: ∠OKA = 107.5 °; ∠AKB = 72.5 °; ∠OBA = 65 °.

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