ABCD is a trapezium in which AB II CD. If angle ADC = twice angle ABC, AD = a cm and CD = b cm, then find the length of AB.

Since the adjacent sides of trapezium are supplementry,

ADC + ABC= 180°.

3ABC = 180°

Also ABC =60°, BAD=60° And ADC=120° and CBA = 120°.

Draw a perpendicular DM bisecting AB, then triangle ADM is formed with DAM =60°. cos60 = 1/2= AM/AD

Hence AM/a=1/2 and AM=a/2.

Similarly the other side perpendicular CN is drawn and cos60 BN = a/2.

The remaining part is same as `b`.

AB = a/2 + a/2 + b = (a+b).