MHB Angle RAB in Triangle PQR: 84°, 78°, 48°, 63°

[anemone]

In a triangle $PQR$, $\angle P=84^{\circ}$, $\angle R=78^{\circ}$. Points $A$ and $B$ are on $PQ$ and $QR$ so that $\angle PRA=48^{\circ}$ and $\angle RPE=63^{\circ}$.

What is the measure of $\angle RAB$?

 

[Albert1]

In a triangle $PQR$, $\angle P=84^{\circ}$, $\angle R=78^{\circ}$. Points $A$ and $B$ are on $PQ$ and $QR$ so that $\angle PRA=48^{\circ}$ and $\angle RPB=63^{\circ}$.

What is the measure of $\angle RAB$?

My solution:

Spoiler

View attachment 3435

 

[johng1]

I couldn't quite follow Albert's solution.

Spoiler

The following also uses the circumcircle of triangle RAB. Also it's written as a slight generalization. It is critical that angle ARQ is 30 degrees. The original problem has $\theta$ equal to 21. So angle RAB is 81.

 

[Albert1]

johng:
Can you follow my solution now ?

Spoiler

View attachment 3439

 

[johng1]

Albert,
I understand perfectly if quadrilateral PDAB is cyclic. This was my original problem, and I still can't see why it's true.

 

[Albert1]

Albert,
I understand perfectly if quadrilateral PDAB is cyclic. This was my original problem, and I still can't see why it's true.

RDAB is cyclic not PDAB

 

[johng1]

Albert,
Sorry, it was a typo. My problem is: why is quadrilateral RDAB cycliC? I can prove this only by knowing the answer.

 

[Albert1]

cyclic-quadrilateral-and-its-properties:
Cyclic Quadrilateral and its Properties | TutorNext.com#

 

[anemone]

I couldn't quite follow Albert's solution.

Spoiler

The following also uses the circumcircle of triangle RAB. Also it's written as a slight generalization. It is critical that angle ARQ is 30 degrees. The original problem has $\theta$ equal to 21. So angle RAB is 81.

Thanks, johng for participating and your solution is correct, well done!:)

cyclic-quadrilateral-and-its-properties:
Cyclic Quadrilateral and its Properties | TutorNext.com#

Hi Albert,

First, thanks for participating in this challenge problem of mine.:)

But, looking at your solution, if you mean to verify that the specific quadrilateral RDAB is a cyclic quadrilateral by first showing the sum of the opposite angles in it are supplementary, then I am unable to follow it.

 

[Albert1]

johng:
Can you follow my solution now ?

Spoiler

View attachment 3439

anemone:
If you are still unable to follow it,I will use the following method (forget the circle)
$x=\angle DRA=30^o=\angle ARB$
$\angle ABQ=x+y=111^o=63+18+x=63^o+48^o$
$\angle RAB=111-30=81^o=y$