We know∠BCD = 43°and∠BAE = 62°

And In∆AEC we get from angle sum property of triangle∠CAE +∠CEA +∠ACE = 180°Here∠BCD =∠ACE = 43°( Same angles )And∠BAE =∠CAE = 62° ( Same angles )And given∠CEA = d , Now we substitute values , we get

62°+ d + 43° = 180°SO,d = 75° ( Ans )

Quadrilateral AEDB is a cyclic quadrilateral , And we know in cyclic quadrilateral opposite angles are supplementary , So

∠AED +∠DBA = 180°, From diagram we substitute values , we get

d + a = 180°So,75° + a = 180° ( As we calculate d = 75°)

a = 105° ( Ans )AndAnd In∆BAF we get from angle sum property of triangle∠BAF +∠AFB +∠FBA = 180°Here∠BAE =∠BAF = 62°( Same angles )And∠FBA = a = 105° ( as we calculate )And given∠AFB = b , Now we substitute values , we get

62°+ b + 105° = 180°SO,b = 13° ( Ans )

And∠DEA and∠DEF are supplementary , as they are linear pair angles , So

∠DEA +∠DEF = 180° , Here∠DEA = d = 75° , So

75° +∠DEF = 180°

∠DEF = 105°In∆DEF we get from angle sum property of triangle∠DEF +∠EFD +∠FDE = 180°Here∠EFD = b = 13°( as we calculate )And∠DEF = 105° ( as we calculate )And given∠FDE = c , Now we substitute values , we get

105°+ 13°+ c = 180°SO,c = 62°