Apologies for the omission of sinBAD and sinCAD. You're right, it was an error. I'm working on the answer again. I apologise for the confusion. My original answer is being revised. I hope I can get it right this time! See below:
It took me a while to interpret the diagram, but I came up with the following picture (finger-drawn on my iPad) (not to scale).
The tensions in the wire labelled T1 and T2 have to counterbalance the weight. BA+AC=15m and T1 and T2 are directly proportional to the lengths of AB and AC. The perpendicular distance Y is the distance AD between A and the 10m horizontal reference line EC at the top of the picture.
We know that, since there is no net horizontal force, T1sinBAD=T2sinCAD, so ABsinBAD=(15-AB)sinCAD; ED+DC=10, so, since ED=ABsinBAD=DC=(15-AB)sinCAD, ABsinBAD=5=(15-AB)sinCAD. sinBAD=5/AB, sinCAD=5/(15-AB). Therefore, cosCAD=sqrt(1-25/(15-AB)^2), cosBAD=sqrt(1-25/AB^2).
Y=AD=2+ABcosBAD=(15-AB)cosCAD
ABcosBAD=sqrt(AB^2-25) and (15-AB)cosCAD=sqrt((15-AB)^2-25).
Y=2+sqrt(AB^2-25)=sqrt((15-AB)^2-25); the angles are not needed; we just need to find AB;
squaring both sides: 4+4sqrt(AB^2-25)+AB^2-25=(15-AB)^2-25=225-30AB+AB^2-25
4sqrt(AB^2-25)=221-30AB, because underlined terms cancel out;
squaring again: 16(AB^2-25)=48841-13260AB+900AB^2;
884AB^2-13260AB+49241=0, from which AB=8.23994 or 6.76006. These values are in fact AB and AC, adding up to 15. But only AB=6.76006 satisfies the equation for Y above, so Y=6.54955m (2+sqrt(AB^2-25) or sqrt((15-AB)^2-25)). Y=6.55 metres to 2 dec places.