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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 12, 2024
 #1
avatar+128796 
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                                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

 

cool cool cool 

 Mar 12, 2024

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