In the picture above, $ABCD$ is a square, $ BQ = QC = 1$ and $AQ = \sqrt{3}$. Find $x+y+z$.
I can see that $BQC$ is isosceles then $\frac{\pi}{2} - y + z + x = \pi$, so $y+x+z = \frac{\pi}{2}+2y$ and by cosine law, $l = \sqrt{4 - 2\sqrt{3}\cos(z)}$, where $l$ is the side of the square. Then $l$ depends on the value of $z$, which I can't see how would be uniquely determined with given informations. How can I solve this problem, if it is solvable?
Thanks.
Rotate $C$ around $B$ for $90^{\circ}$ in to $A$ and let $Q'$ be image of $Q$ under this rotation. Then triangle $Q'BQ$ is right isosceles triangle so $QQ' = \sqrt{2}$ and sice $AQ'=1$ we see that triangle $Q'AQ$ is right at $Q'$. So $$\angle AQ'B = 90^{\circ}+45^{\circ} =135^{\circ}$$
Can you finish?