+0  
 
0
47
2
avatar+464 

this question is based on Unit Cricle Trigonometry

 

shreyas1  Oct 11, 2018
 #1
avatar+3175 
+1

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

Rom  Oct 11, 2018
 #2
avatar+464 
+1

Thank you rom once again

shreyas1  Oct 11, 2018

14 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.