choutowne

Thank you very much!! 

Your detailed explanation made it much easier to understand the answer.

And sorry I didn't clarify, A must be a single-digit positive integer, so the only value possible in that set is 6, which is the answer :)

Thank you so much!

Your explanation helped a lot.

Nevermind, I figured it out with symmetry! If anyone's wondering, it's \(\boxed{96}\)

Nevermind, thanks, I got the answer \(\boxed{\frac{16}{7}}\)! Thanks for your help!

Thanks for the answer, it helped me a lot!

5sqrt(21)/41 wasn't correct, but your response helped me figure out what to do next.

I followed your solution and got mostly the same results except that I got 40/41 as cos(PAQ) instead of 34/41.

I plugged that into the relationship between the sine and cosine of PAQ and got 9/41 as the final answer, which turned out to be correct. 

\( sin(PAQ) = \boxed{9/41}\)

You just copied CPhill's answer to my previous question...

Thank you so much! I forgot to mention but I believe the answer should be in the form of a common fraction. However, I'm not sure how I should convert it. (It's a program where one inputs the answer and doesn't take sin/cos etc. in the expression)  

Sorry, I'm a little lost. There are multiple points in your answer which confuse me. 
Why is XY = XZ = radius (I'm assuming R is radius) because X lies on the perpendicular bisectors of all three sides of the triangle? Points Y and Z lie on a circle centered on O, so OY and OZ would be the radius, unless you're referring to another circle? 

What is the inscribed circle? 

How did you get 48/(2 * sqrt(3)) = 12sqrt(3)?

Sorry, and thanks.