# math

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. Call
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

posted by Steve

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