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.
C
45
A P Q B
arccos (2/3) = ACP ≈ 48.18°
Impossible ACB = 90
But ACP +QCP = 45 + 48.18 = 93.18 which is > 90
+760 
