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Right Triangle Trigonometry 

 

 
shreyas1  Oct 11, 2018
 #1
avatar+2389 
+2

 

From the diagram about you can see that

 

\(|PX|=\sqrt{5^2+12^2} = 13 \\ \cos\left(\dfrac 1 2 \angle PXS\right)=\dfrac{12}{13} \\ \sin\left(\dfrac 1 2 \angle PXS\right) = \sqrt{1-\dfrac{12}{13}} = \dfrac{5}{13} \\ \cos(2x) = \cos^2(x) - \sin^2(x) \\ \cos\left(\angle PXS\right) = \cos\left(2\cdot \dfrac 1 2 \angle PXS\right) = \dfrac{144}{169}-\dfrac{25}{169} = \dfrac{119}{169}\)

 
Rom  Oct 11, 2018
 #2
avatar+464 
+1

Thank you Rom

 
shreyas1  Oct 11, 2018

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