Asked by RhUaNg
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.
Answers
Answered by
bee
I have the same question! will someone please answer it by today
Answered by
hi
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Answered by
aasdf
need this too please help!
Answered by
Anonymous
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.
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