MATH_URGENT

Point X is on side \overline{AC} of \triangle ABC such that \angle AXB =\angle ABX, and \angle ABC - \angle ACB = 39 degrees. Find \angle XBC in degrees.

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  1. I have the same question! will someone please answer it by today

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  2. i need it too

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  3. need this too please help!

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  4. Please remember that you can't just ask for the answers, you need to provide some steps you already got and what steps you can't solve. But here, I will just show you guys the steps and you can get the answer easily.

    Steps:
    Since angle AXB = angle ABX, we have angle XBC = angle ABC - angle ABX = angle ABC - angle AXB. angle AXB is an exterior angle of triangle XBC, so angle AXB = angle C + angle XBC. Therefore, we have angle XBC = angle ABC - (angle C + angle XBC), so 2 angle XBC = angle ABC - angle C = 39°. Thus, angle XBC = ?.

    ? is the part where you need to solve by yourself.

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