The angles in triangle ABC satisfy 6sin∠A=3√3sin∠B=2√2sin∠C. If sin2∠A=a/b, where a and b are coprime positive integers, what is the value of a+b?

This is a brilliant problem

I think the question needs to be rephrased as follows:

The angles in triangle ABC satisfy 6sin∠A=3√3sin∠B=2√2sin∠C. If sin2∠A, calculated correctly to 2 decimal places,= a/b where a and b are co-prime positive integers, what is the value of (a+b)?

Let 6sin(A)=3√3sin(B)=2√2sin(C) = k
This gives sin(A)= k/6,
sin(B)=k/(3√3),cos(B)=(1-k^2/27)^(1/2)= (27-k^2)^(1/2)/(3√3)
sin(C)=k/(2√2),cos(C)=(1-k^2/8)^(1/2)=(8-k^2)^(1/2)/(2√2)
Cos(A)=(1-k^2/36)^(1/2)
Sin(2A) = (1/3)k(1-k^2/36)^1/2
= k(36-k^2)^(1/2)/18
Sin(A)=sin(B+C)=sin(B)cos(C)+sin(C)cos(B) or
k/6=(k/(3√3))(1-k^2/8)^(1/2)+(1-k^2/27)^(1/2)(k/(2√2))
or (6)^(1/2) = (8-k^2)^(1/2) + (27-k^2)^(1/2) which gives k^2 = 23/24
And this further yields: sin(A)= 29/(6√24) & sin(2A)=(29√23)/(24*18)~ = 0.32 correct to 2 decimal places = 8/25 so (a+b) = 33