Let $ABCD$ be a square. Let $P$ be a point outside square $ABCD$ such that triangle $ABP$ is equilateral. Find $\angle BPD$ in degrees.
Note that angle PAD = 150
AD = AP
So
angles ADP and APD are equal
APD = (180 -150) /2 = 15°
Therefore angle BPD = 60 -15 = 45°
+635 
