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this question is based on Unit Cricle Trigonometry

 

 Oct 11, 2018
 #1
avatar+4799 
+1

\(\text{Probably the clearest way to do this is to note that}\\ \cos(\angle RPS)=\cos\left(\pi - \angle RPQ\right)=-\cos(\angle RPQ)\\ \cos(\angle RPQ)=\sqrt{1 -\sin^2(\angle RPQ)} = \sqrt{1-\frac{49}{625}} \\ \cos(\angle RPQ) = \sqrt{\dfrac{576}{625}}=\dfrac{24}{25},\text{ and thus}\\ \cos(\angle RPS)=-\dfrac{24}{25} \)

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 Oct 11, 2018
 #2
avatar+464 
+1

Thank you rom once again

 Oct 11, 2018

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