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.
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