In triangle $ABC,$ $AC = BC$ and $\angle ACB = 90^\circ.$ Points $P$ and $Q$ are on $\overline{AB}$ such that $P$ is between $A$ and $Q$ and $\angle QCP = 45^\circ.$ If cos ACP = 2/3, then find cos BCQ.
Impossible
To see why
ACB = 90°
QCP = 45°
If the cos ACP = 2/3....then arcos (2/3) ≈ 48.2°
But this means that QCP + ACP = 45 + 48.2 = 93.2° which would be > ACB
+704 
