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As shown in the figure, ⊙ o is the circumscribed circle of ⊙ ABC, ﹤ BAC = 55 ° and the degree of ﹤ OBC is () A. 25° B. 35° C. 55° D. 70°
⊙ o is the circumcircle of ⊙ ABC, ∵ BAC = 55 °,
∴∠BOC=2∠BAC=2×55°=110°，
∵OB=OC，
∴∠OBC=180°−∠BOC
2=180°−110°
2=35°．
Therefore, B
As shown in the figure, in ⊙ o, ⊙ B = 50 °, C = 20 °, find the size of ⊙ BOC
Connect OA,
∵AO=BO=CO，
Both △ OAB and △ OAC are isosceles triangles,
∴∠BAO=∠B=50°，
∠CAO=∠C=20°
∴∠BAC=70°，
∴∠BOC=2∠BAC=140°．
As shown in the figure, points a, B and C are all on the circle O. if the angle ABC is equal to 70 degrees and the angle ACB is equal to 50 degrees, find the degree of the angle BOC
Angle BOC = 2 times of angle a = (180 ° - 70 ° - 50 °) × 2 = 120 °
As shown in the figure, in the circle O, we know BC = AC, ∠ ABO = 50 ° and find the degree of ∠ BOC Come on, I'm in a hurry, Gogo
Connect OC, extend Bo to intersect the circle to D, the angle ABO = 50 degrees, then the angle AOD = 100 degrees (because the circle center angle = 2 times the circumference angle), the angle AOB = 80 degrees, because BC = AC, so the triangle OBC is equal to the triangle OAC,
So the angle BOC = 1 / 2, the angle AOB = 40 degrees
As shown in the figure, in the circle O, AB is the diameter, BC = CD = De, ∠ BOC = 50 ° and find the degree of ∠ AOE
Because BC = CD = De,
So the angle BOC = angle cod = angle DOE = 50 degrees
So the angle BOE is 150 degrees
Because boa is 180 degrees
So the angle AOE is 30 degrees
As shown in the figure, the point O is the center of the tangent circle of △ ABC. If ∠ BAC = 80 °, then the degree of ∠ BOC is a.130 ° b.100 ° c.50 ° d.65 °
Because the circle O is the inscribed circle of △ ABC, so o is the heart of the triangle, that is, the intersection of the bisectors of the three angles of the triangle
Because ∠ BAC = 80 °, so ∠ B + ∠ C is equal to 100 degrees,
So ∠ OCB + ∠ OBC = 50 degrees, so ∠ BOC = 180 degrees - 50 degrees = 130 degrees
Option a
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