Given, ∠BAC = 70°. Now, in △ ABC using angle sum property of triangle, we have ∠ABC + ∠ACB + ∠BAC = 180° ⇒∠ABC + ∠ACB + 70° = 180° ⇒ ∠ABC + ∠ACB = 180° - 70° = 110° ⇒ 2∠OBC + 2∠OCB = 110° [Given BO and CO are bisectors of ∠ABC and ∠ACB respectively. So, ∠ABC = 2∠OBC and ∠ACB = 2∠OCB] ⇒ ∠OBC + ∠OCB = 55° In △ BOC using angle sum property of triangle, we have ∠OBC + ∠OCB + ∠BOC = 180° ⇒ 55° + ∠BOC = 180° ⇒ ∠BOC = 180° - 55° = 125° Posted by Ankush Jain(MeritNation Expert)on 13/7/12 This conversation is already closed by Expert

Given, ∠BAC = 70°. Now, in △ ABC using angle sum property of triangle, we have ∠ABC + ∠ACB + ∠BAC = 180° ⇒∠ABC + ∠ACB + 70° = 180° ⇒ ∠ABC + ∠ACB = 180° - 70° = 110° ⇒ 2∠OBC + 2∠OCB = 110° [Given BO and CO are bisectors of ∠ABC and ∠ACB respectively. So, ∠ABC = 2∠OBC and ∠ACB = 2∠OCB] ⇒ ∠OBC + ∠OCB = 55° In △ BOC using angle sum property of triangle, we have ∠OBC + ∠OCB + ∠BOC = 180° ⇒ 55° + ∠BOC = 180° ⇒ ∠BOC = 180° - 55° = 125° Posted by Ankush Jain(MeritNation Expert)on 13/7/12 This conversation is already closed by Expert

Given, ∠BAC = 70°. Now, in △ ABC using angle sum property of triangle, we have ∠ABC + ∠ACB + ∠BAC = 180° ⇒∠ABC + ∠ACB + 70° = 180° ⇒ ∠ABC + ∠ACB = 180° - 70° = 110° ⇒ 2∠OBC + 2∠OCB = 110° [Given BO and CO are bisectors of ∠ABC and ∠ACB respectively. So, ∠ABC = 2∠OBC and ∠ACB = 2∠OCB] ⇒ ∠OBC + ∠OCB = 55° In △ BOC using angle sum property of triangle, we have ∠OBC + ∠OCB + ∠BOC = 180° ⇒ 55° + ∠BOC = 180° ⇒ ∠BOC = 180° - 55° = 125° Posted by Ankush Jain(MeritNation Expert)on 13/7/12 This conversation is already closed by Expert