Question

In the triangle ABC, the bisectors of the
exterior angles B and C meet at 0.
Given angle ABC = 80° and angle BAC = 50°.
Find angle BOC
plz give answer only if you know and show each step plz​

Answer 1

Answer:

In △ABC,

∠ABC+∠BAC+∠ACB=180

60+70+∠ACB=180

∠ACB=50

∘

ext.∠B=180−∠ABC

2∠OBC=180−70 (BO bisects ext.∠B)

∠OBC=55

∘

Similarly, ∠OCB=65

∘

Now, In △OBC,

∠OCB+∠OBC+∠BOC=180

65+55+∠BOC=180

∠BOC=180−1/20

∠BOC=60

∘

this is the answer.