Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
Point X is on side \overline{AC} of \triangle ABC such that \angle AXB =\angle ABX, and \angle ABC - \angle ACB = 39 degrees. F...Asked by HELPPPPP
Point $P$ is on side $\overline{AC}$ of triangle $ABC$ such that $\angle APB =\angle ABP$, and $\angle ABC - \angle ACB = 39^\circ$. Find $\angle PBC$ in degrees.
Answers
Answered by
Reiny
How about just typing:
Point P is on side AC of triangle ABC such that angle APB =angle ABP, and angle ABC - angle ACB = 39°. Find angle PBC in degrees.
let angle APB = angle ABP = x
let angle PBC = y, let angle BCP = k
given:
x+y - k = 39
k = x+y-39
In triangle PBC, the angles add up to 180°
y + 180-x + k = 180
y + 180 - x + x+y-39 = 180
2y = 39
y = 39/2 = 19.5°
check my arithmetic
Point P is on side AC of triangle ABC such that angle APB =angle ABP, and angle ABC - angle ACB = 39°. Find angle PBC in degrees.
let angle APB = angle ABP = x
let angle PBC = y, let angle BCP = k
given:
x+y - k = 39
k = x+y-39
In triangle PBC, the angles add up to 180°
y + 180-x + k = 180
y + 180 - x + x+y-39 = 180
2y = 39
y = 39/2 = 19.5°
check my arithmetic
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.