Question

已知:如图①,在△ABC中,∠ABC和∠ACB的外角平分线交于O

Answer 1

(1) ∠ABC+∠ACB = 180 - ∠A

∠ABC 和 ∠ACB的外角和为 180 - ∠ABC+ 180 - ∠ACB = 360 - (180 - ∠A) = 180 + ∠A

∠BOC = 180 - 1/2(180+∠A) = 90 - 1/2∠A = 90 - 20 = 70

(2) 这个角的度数哪能算出来...