Let $ABCD$ be a square. Let $P$ be a point outside square $ABCD$ such that triangle $ABP$ is equilateral. Find $\angle BPD$ in degrees.
P
A B
C D
Angle ABP = 60
Angle ABD = 90
Angle PBD = 60 + 90 = 150
In triangle PBD , BP = BD
So angle BDP = angle BPD
So angle BPD = (180 - 150) / 2 = 30 / 2 = 15°
+1248 
