In triangle ABC the bisector of angle A meet BC at point D and AB = CD and B= 2 angle C Find Angle BAC numerically
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Answer:
In △BOC,
∠BOC+∠OBC+∠OCB=180 (OB and OC bisect ∠B and ∠C respectively)
∠BOC+
2
1
∠B+
2
1
∠C=180
∠BOC=180−
2
1
(∠B+∠C)
∠BOC=180−
2
1
(180−∠A)
∠BOC=180−
2
1
(180−70)
∠BOC=180−90+35
∠BOC=125
∘
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