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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.

 Mar 2, 2024
 #1
avatar+128794 
+1

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

 

cool cool cool

 Mar 3, 2024

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