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

blackpanther Nov 19, 2023

#1**0 **

**angle PCD = 75 ^{o}**

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 60^{o} of the

equilateral triangle and the 90^{o} of the square, for a total of 150^{o}.

The sum of the interior angles of any triangle is 180^{o} and so that

leaves 30^{o} for the two small angles to share, namely 15^{o} each.

At corner C, subtract that 15^{o} from the 90^{o} of the square, leaving 75^{o}.

_{.}

Bosco Nov 20, 2023