In the triangle ABC, the angle a = 62 degrees, EO and fo are the vertical bisectors of AB and AC respectively and intersect at point O. calculate the degree of the angle BOC

In the triangle ABC, the angle a = 62 degrees, EO and fo are the vertical bisectors of AB and AC respectively and intersect at point O. calculate the degree of the angle BOC

∠A=62°,∠ABC+∠ACB=180-62=118°
EO and fo divide AB, AC and connect Ao vertically,
Then ∠ ABO = ∠ Bao, ∠ Cao = ∠ OCA
∠BAO+∠CAO=∠A=62°,
Then ∠ ABO + ∠ OCA = 62 °, ∠ OBC + ∠ OCB = 118 ° - 62 ° = 56 °
So ∠ BOC = 180 ° - 56 ° = 124 °

In the acute triangle ABC, a = 50 degrees, AC.BC The vertical bisectors on both sides intersect at O, and calculate the degree of ∠ BOC

Solution 1: ∵ AC.BC Solution 2: connect OA. ∵ vertical bisector ∵ OA = OC, = ob ∵ OAC = ∵ OCA, ∵ OAB = ∵ oba ∵ BAC = ∵ OAB + ∵ OAC = 50 °

In the acute angle △ ABC, a = 50 degree AC.BC If the vertical lines on both sides intersect at the point O, then the degree of ∠ BOC is -?

AC.BC If the vertical lines on both sides intersect at point O, then OA = ob = OC ∠ OAC = ∠ OCA ∠ OAB = ∠ oba
∠OBC=∠OCB ∠OBA+∠OCA=∠A=50° ∠BOC=∠A+∠OBA+∠OCA=2∠A=100°
The degree of BOC is 100 degrees