Sue VanHattum · Mar 2

Are we to assume that the two sides of the angle are the same length?

Catriona Agg · Mar 2

I like your isosceles triangle solution; I wouldn’t have thought of that approach.

Tim Corica · Mar 6

Here's a cheap way: The prob is underspecified - we can't determine the orientation of the big SQ. Let bottom rt of the bottom pink sq be (0,0). We know (0,4) lies on SQ. Rotate SQ c.w. until a vertex is at (0,4). This satisfies the givens. The angle is 45. Cheap, but correct (I think!).

Dr Rick · Mar 2

Very nice! Consider the circle whose diameter is the vertical line of four square sides. The three points defining the angle are all on this circle so it's equal to the angle with the same base at the top of the diameter, which is 45°.