Verified answer

sinx=12/13

sin2x=2sinxcosx

cosx=sqrt(1-sin^2x)

cosx=5/13

hence sin2x=120/169

Sin(2β) = Sin (β+β) = Sinβ Cosβ + Cosβ Sinβ = 2Sinβ Cosβ

Now Sinβ = 12/13

=> [ 1 - (Cosβ)^2 ]^(1/2) = 12/13

=> [ 1 - (Cosβ)^2 ] = (12/13)^2 = 144/169

=> (Cosβ)^2 = 1 - 144/169 = (169 - 144) /169 = 25/169

=> Cosβ = (25/169)^ (1/2) = 5/13

So, Sin(2β) = 2Sinβ Cosβ

=>Sin(2β) = 2 * (12/13) * ( 5/13) = 120/169 ..... (ANSWER)

the answer has two possibilities...sin ß = 12 / 13 ---> cos ß = 5 / 13 OR - 5 / 13

and sin 2ß = 2 sin ß cos ß ...you finish...note : one set is not adequate