It is known that the side length of square ABCD is 2, and there are a little p and q on AB and AD. If the perimeter of △APQ is 2, find.

∠∠∠ ∠PCQ.

Solution: analyze it first. The diagram in your topic is not very accurate, because neither Q nor P can cross the midpoint between AD and AB, otherwise a triangle cannot be formed.

Besides, there is more than one such triangle. If each triangle has a different number of angles, then this topic is meaningless. So you can use a special triangle to find this angle, and it is best to use a right isosceles triangle. Let's start asking.

As shown in the figure, make a right-angled isosceles triangle APQ, take the midpoint O of its hypotenuse and connect CO, then CO is the diagonal of the square; Then take the midpoint r and s of AD and AB. Because the perimeter of △APQ is 2 and the side length of a square is 2, PQ must be equal to QR+PS, so there are, PQ=(√2)AQ, OQ=QR=PQ/2=(√2)AQ/2, AQ+QR= 1.

Because AC = 2 √ 2 and AO = OQ = (√ 2- 1)/2, OC=AC-AO=(3√2- 1)/2, in Rt△OCQ.

Tan (≈OCQ)= OQ/OC =(5-√2)/ 17.