math
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.?

Steve
Call
AXB=ABX = a
ABC = x
ACB = XCB = x39
We want XBC = xa
Then we have
2a+A = 180
xa + x39 + 180a = 180
2x  2a = 39
xa = 19.5
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