Geometry problem no.-5

[Sujeet Kumar]

Here is geometry problem no.-5 :-
Q.In a triangle ABC, where angle B is obtuse, point P is taken so that angle ABP=(180-3x) degree,
angle PBC=x degree, angle PCB=30 degree and angle PCA=(2x-60) degree.Find angle BAP.

[Sujeet Kumar]

Here is the solution :-

Q.In a triangle ABC, where angle B is obtuse, point P is taken so that angle ABP=(180-3x) degree,
angle PBC=x degree, angle PCB=30 degree and angle PCA=(2x-60) degree.Find angle BAP.

Sol.-Angle BAC=180-(180-3x+x+30+2x-60)=180-150=30 degree.
Let O be the circumcenter of the triangle ABC.
Angle AOC(major)=2*(180-3x+x)=(360-4x) degree.
Angle AOC(minor)=360-(360-4x)=4x degree.
Triangle PCB will be congruent to PCO.
Angle POC=angle PBC=x degree.
Angle AOP=angle AOC- angle POC=4x-x=3x degree.
Considering quadrilateral ABPO,
Angle ABP+angle AOP=(180-3x)+3x=180 degree.
So, quadrilateral ABPO is cyclic.
Hence, angle BAP=angle BOP=angle BOC-angle POC=(60-x) degree.