In triangle ABC, AC=BC and ∠ACB=90∘. Points P and Q are on ¯AB such that P is between A and Q and ∠QCP=45∘. 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