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.
4 answers
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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.
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.