# 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 °.