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In the acute triangle ABC, if the angle a = 50 degrees and ab is greater than BC, the value range of angle B is in the range Wrong. AB and BC are sides
AB is greater than BC
So angle c is greater than angle a by more than 50 degrees
So the range of angle B is greater than 0 degrees and less than 80 degrees
And because it's an acute triangle
So the maximum angle c is infinitely close to 90 degrees, so the minimum angle B is 40 degrees and the maximum angle B is 80 degrees
Final answer (40 ° and 80 °)
It is known that O is the intersection point of the vertical bisector of ABC three sides of an acute triangle. The angle OCB = 90 degree angle BAC is proved
Prove: A, B, C are three points on the circle with the circle O as the center. Then lead the long line co to intersect on the circle and mark it as point E, then CE is the diameter of the circle O, connect be (according to the above words, you can draw the picture, I will not draw), then angle e is equal to angle a (because they correspond to the common arc), because CE is the straight arc of the circle
In the triangle ABC, if the angle a is equal to 70 degrees and O is the intersection of the vertical bisectors of the sides AB and BC, then the angle BOC is equal to 70 degrees
Connecting OA, ob, OC & nbsp; ∵ o is the intersection of the vertical bisectors of AB and BC ∵ ob = OA = OC & nbsp; & nbsp; ∵ OAB + ∵ OAC = ∵ a = 70 °∵ OA = ob, OA = OC ∵ oba + ∵ OCA = ∵ OAB + ∵ OAC = 70 °∵ triangle inner angle and 180 °
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