posted by G

Point X is on side of line AC of triangle ABC such that <AXB = <ABX, and <ABC - <ACB = 39 degrees. Find <XBC in degrees.?

  1. Steve

    AXB=ABX = a
    ABC = x
    ACB = XCB = x-39

    We want XBC = x-a

    Then we have

    2a+A = 180
    x-a + x-39 + 180-a = 180
    2x - 2a = 39
    x-a = 19.5

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