Number one geometry problem

[Sujeet Kumar]

Following is geometry problem no.-1 :-
Q.In a triangle ABC, point P is taken so that angle ABP=x degree=angle ACP and angle ABC=30 degree=angle PCB.Find angle PAC,

[Mary Krimmel]

This is an interesting problem. I cannot solve it, but I find the "answer" to be (60-2x) degrees, 0<x<30.

[Sujeet Kumar]

> Following is geometry problem no.-1 :-
> Q.In a triangle ABC, point P is taken so that angle
> ABP=x degree=angle ACP and angle ABC=30 degree=angle

> PCB.Find angle PAC.

Here is the solution :-
Q.In a triangle ABC, point P is taken so that angle ABP=x degree=angle ACP and angle ABC=30 degree=angle PCB.Find angle PAC.
Sol.-Construct an equilateral triangle MBC so that M is on the same side of BC as of A.
Triangle ABC is congruent to triangle ABM.
Angle BMA=(30+x) degree.
Triangle PCB is congruent to triangle PCM.
Angle PMC=(30-x) degree.
Suppose angle BMP> angle BMA
Angle BMA + angle AMP+ angle PMC=60 degree=(30-x) + (30+x)=angle BMA+ angle PMC
=> angle BMA+angle AMP= angle BMA
=> angle AMP= 0 degree ie., A , M and P are collinear.
Angle APC=angle BPC=180-(30+30 -x)=(120+x) degree.
Angle PAC=180-(angle PCA+ angle APC)=180-(x+120+x)=(60-2x) degree.