there is a trapezium ABCD.AB is parallel to CD.Angle A is twice of angle C.If AD=a and AB=b.Then what is the value of DC in terms of a and b?
DB' = BD' = AB - AD' = b - a cos(2 alpha)
CD = DB' + B'C = DB' + BB' / tan(alpha) = DB' + DD' / tan(alpha)
CD = b - a cos(2 alpha) + a sin(2 alpha) / tan(alpha)
CD = b - a (2 cosē(alpha) - 1) + a 2 sin(alpha) cos(alpha) cos(alpha) / sin(alpha)
CD = a + b
Of course you have also to consider the case where D' lies at the right of B