Re: Nothing special geometry problem

[Sujeet Kumar]

> There is nothing special about following geometry
> problem-
> Q.In a triangle ABC, point P is taken so that angle
> ABP=20 degree, angle PBC=40 degree, angle PCB=30
> degree and angle PCA=10 degree.Find angle PAC.
>
> **I am missing Mary's and Narsimhan's responses too.**

Here is the solution-
Q.In a triangle ABC, point P is taken so that angle ABP=20 degree, angle PBC=40 degree, angle PCB=30 degree and angle PCA=10 degree.Find angle PAC.
Sol.-Take a point M on AB , so that A is in the middle of M and B and triangle MBC is equilateral.
Angle BAC=180-(20+40+30+10)=180-100=80 degree.
Let BP extended meet MC at K.
Angle BKC=180-(40+60)=80 degree.
Angle BAC=angle BKC
So, quadrilateral AKCB is cyclic.
Angle KAC=angle KBC=40 degree.
Triangle PCB will be congruent to triangle PCM.
So, angle PMC=angle PBC=40 degree.
Angle PKM=180-angle BKC=180-80=100 degree.
Considering triangle MKP, angle MPK=180-(40+100)=40 degree.
So, MK=PK
Angle AKB=angle ACB=40 degree=angle AKP.
Angle AKM=180-(angle BKC+angle AKP)=180-(80+40)=60 degree.
So, triangle AMK will be equilateral.
So, MK=AK=PK
So, triangle AKP is isosceles.
Angle AKP=40 degree.
So, angle PAK=70 degree.
Hence, angle PAC=angle PAK-angle KAC=70-40=30 degree.