(Apologies for the vague title, I could not figure out how to describe it adequately within the word limit)
I came across this problem in an online forum a few months ago, with no answers under it. The diagram looks like this:
We have a square and another right angle at its lower left vertex, and the question is to find the missing angle formed by the intersection of the diagonal of this square, and another line segment of equivalent length. This geometry problem is quite interesting and unique compared to what I’ve posted before, there is not a lot of information given.
I will post my solution below as an answer. Please let me know if my answer is correct, and if there are any problems in my approach. Also, please post your own answers as well, preferably using different methods in order to ensure my answer is accurate.
Best Answer
Consider the diagram
Since $\angle PAD$ and $\angle QAB$ are both complementary to $\angle DAQ$, $$ \angle PAD=\angle QAB\tag1 $$ Thus, by SAS, we have $$ \triangle PAD\simeq\triangle QAB\tag2 $$ Note that $\angle PDA=\angle QBA=\frac\pi4$. Therefore, $$ \begin{align} \angle PDQ&=\frac\pi2\tag{3a}\\ \angle PDC&=\frac{3\pi}4\tag{3b} \end{align} $$ The Law of Sines says that $$ \begin{align} \sin(\angle DPC) &=\frac{DC}{PC}\,\sin(\angle PDC)\tag{4a}\\[3pt] &=\frac1{\sqrt2}\frac1{\sqrt2}\tag{4b}\\ &=\frac12\tag{4c} \end{align} $$ Since $\angle DPC=\frac\pi6$ and $\triangle RDP$ is a right triangle, we get that $\angle DRP=\frac\pi3$. Therefore, since vertical angles are equal $$ \angle BRC=\frac\pi3\tag5 $$