Mark the point P inside a square ABCD that makes the triangle CDP equilateral. Calculate the size of angle PAD?
I got around 77. (I used a protractor)
Could I get some hints on how to get the angle?
First you should have a clear idea of what the thing you've drawn looks like:
I've taken the liberty to highlight the angles I find important for solving this problem.
Then you should take note that all angles within triangle DCP are the same, and that they add up to 180 degrees.
Also you should notice that $\displaystyle \angle{APD}=\angle{PAD}~\mbox{since}~\triangle{APD }~\mbox{is isosceles}$
And lastly, you should notice that $\displaystyle \angle{ADP}+\angle{PDC}=\angle{ADC}=90^\circ$
One last point: never use a protractor(or any other tool for that matter) to solve a geometrical problem of any kind, unless the problem explicitly states that you should. The risk that you will end up with a faulty answer(which you actually did in this case) is huge.