Answer:
[tex]Solution,\\<ACB = <CAF = 90^o [BC ~is~ parallel~ to~ CF~ and ~alternate~ angles]\\In~triangle~ABC,\\<ACB +<ABC+<BAC = 180^o [Sum~of~angles~of~triangle~is~180^o]\\or, 90^o + 28^o +<BAC = 180^o\\or, <BAC = 62^o \\Now,\\<BAC+<AED = <BDE [Sum~of~interior~angles~triangle~is~equal~to~the~exterior~angle]\\or, 62^o+3x = 4x+x\\or, 62^o = 5x-3x\\or, 2x = 62^o\\or, x = 31^o[/tex]
