Edexcel M1 Dynamics [ Mixed Ex 3I Q26 (b) ]

I am stuck on M1 Mixed Ex3I, Q26 (b) in the Edexcel textbook. I'll quote the question.

Two particles A and B, of mass m kg and 3 kg respectively, are connected by a light inextensible string string. The particle A is held resting on a smooth fixed plane inclined at 30 degrees to the horizontal. The string passes over a smooth pulley P fixed at the top of the plane. The portion AP of the string lies along a line of greatest slope of the plane and B hangs freely from the pulley, as shown in the figure. The system is released from rest with B at a height of 0.25 m above horizontal ground. Immediately after release, B descends with an acceleration of 2g/5. Given that A does not reach P, calculate

(b) the value of m.

I tried doing this resolving directly upward/north, and I can't seem to get the correct answer (2). I have seen the solution where you can resolve in a north east direction and I understand it, but I just want to know if it is possible to resolve north and still get the answer, as that was what I instinctively turned to.

My workings included (where R is the normal reaction, a is the acceleration and T is worked out as 17.64) :

[Resolving north west]

R = mgcos30+(2g/5m)

[Resolving upwards]

mg = Rcos30+Tcos60+acos60

mg = [mgcos30+(2g/5m)]cos30+(17.64)cos60+(2g/5)(cos(60))

9.8m = m( cos30+ 2g/5) + 10.78

5.01... m = 10.78

m = 2.14999...

So I got close, but I'm pretty sure the solution resolving north east gets m = 2 without any decimals. I know this is a bit more complicated than the other way of resolving, I just want to know if my way of seeing a question would fail in an exam. Thanks.

how u calculated this ?(bold) 2g/5 is the acceleration of both A and B right ?

I was just focusing on A throughout all of this. Sorry for not making this clear, I'll edit the post again!

no i menat your calculatrion of R is wrong its R=mgcos30 no need +2mg/5

Yes I think I made a mistake there since the acceleration is perpendicular, my fault. But even using R = mgcos30, I still don't get the correct answer

draw a free body diagram.
to be frank u dont need the value of R
T-mgsin30=2mg/5
do u get this ?

[Resolving upwards]

mg = Rcos30+Tcos60+acos60

Yes I have seen that solution, but that is resolving in a different direction. I just wanted to know if it was possible to resolve directly upward and still get the correct answer

This equation is also incorrect.

If you resolve vertically, then using ma=F.

ma cos60 = Rcos30 + Tcos60 - mg

Which rearranges to something slightly different to what you had.

Yes I have seen that solution, but that is resolving in a different direction. I just wanted to know if it was possible to resolve directly upward and still get the correct answer

This equation is also incorrect.

If you resolve vertically, then using ma=F.

ma cos60 = Rcos30 + Tcos60 - mg

Which rearranges to something slightly different to what you had.

Yes that makes sense now, thank you. What value for R did you get? I tried using R=mgcos30 but I didn't get the right answer

Edit: I forgot to put the negative on the acceleration since it was in the opposition direction, I got the correct answer now. I apologize, my brain has not been working today