There is a trapezium \(ABCD\) with parallel sides \(AB\) and \(DC\) and height 1. It is known that \(\angle DAB = 6^{\circ}\) and \(\angle ABC = 42^{\circ}\). Point \(P\) is on side \(AB\) such that \(\angle APD = 78^{\circ}\) and \(\angle CPB = 66^{\circ}\).

What is the value of \(AD-BC+DP-CP\)?

×

Problem Loading...

Note Loading...

Set Loading...