Geometry

Question

0

39

1

+750

Let $ABCD$ be a square. Let $P$ be a point outside square $ABCD$ such that triangle $ABP$ is equilateral. Find $\angle BPD$ in degrees.

CPhill · Answer 1 · 2024-06-05T01:32:36

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°