Geometry problem no.-3

[Sujeet Kumar]

Here is geometry problem no.-3 :-
Q.In a triangle ABC, point P is taken so that angle PCB=x degree and angle PBC=30 degree.Find angle BAP.It is given that angle ABC=[135-(x/2)] degree and angle ACB=30 degree.

[Mary Krimmel]

Will you give an overall solution? Again, here's no solution. Angle BAP = x degrees, 0<x<30

[Sujeet Kumar]

After 7 days I will give solution , these problems are just the generalisation of previous problems.If
you see the solution of my previous problems , you can find the solution in just the same way.

Regards

[Sujeet Kumar]

Here is the solution-

Q.In a triangle ABC, point P is taken so that angle PCB=x degree and angle PBC=30 degree.Find angle BAP.It is given that angle ABC=[135-(x/2)] degree and angle ACB=30 degree.

Sol.-Construct an equilateral triangle so that M is on the same side of BC as P.
Triangle PBC will be congruent to triangle PBM.
Angle BMP=angle BCP=x degree.
Angle PMC=(60-x) degree.
Triangle ACB will be congruent to triangle ACM.
Angle AMC=angle ABC=[135-(x/2)] degree.
Angle AMP=angle AMC- angle PMC=[135-(x/2)]-(60-x)=[75+(x/2)] degree.
Angle PBA=[135-(x/2)]-30=[105-(x/2)] degree.
Considering quadrilateral ABPM,
Angle ABP+angle AMP=[105-(x/2)]+[75+(x/2)]=180 degree.
So, quadrilateral ABPM will be cyclic.
Hence , angle BAP=angle BMP=x degree.