In the figure shown, $ABCD$ is a square and $\triangle CDE$ is equilateral. What is the degree measure of $\angle CBE$?
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By the Cosine Law, BE = 1.8794, and
cos angle EBC = (BE^2 + BC^2 - CE^2)/(2*BE*BC) = cos 20, so angle EBC = 20 degrees.
+538 Triangle CBE is isosceles because the square side length is equal to the triangle side length.
TO find the remaining angles, we find the middle angle, and the other two will be half because it is isosceles.
Angle ACB is equal to the squares interior angle plus the interior of the triangle.
90+60=150.
The remaining 2 angles are equal to each other.
So 180-150=30.
30/2=15.
