+1471 Let $ABCD$ be a square. Let $P$ be a point outside square $ABCD$ such that triangle $ABP$ is equilateral. Find $\angle PCD$ in degrees.
+1277
angle PCD = 75o
Grab a scrap of paper and sketch it.
The line from P to C is the base of a isoceles triangle.
Its equal sides are a side of the square and a side of the triangle.
The big angle of the isoceles triangle is the sum of the 60o of the
equilateral triangle and the 90o of the square, for a total of 150o.
The sum of the interior angles of any triangle is 180o and so that
leaves 30o for the two small angles to share, namely 15o each.
At corner C, subtract that 15o from the 90o of the square, leaving 75o.
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