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The terminal ray of angle  β , drawn in standard position, passes through the point (−5, 2sqrt7) .

What is the value of cosβ ?

Enter your answer, in simplest radical form

 

cosβ=  

 Jan 23, 2020
 #1
avatar+8956 
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The terminal ray of angle β , drawn in standard position, passes through the point (−5, 2sqrt7) .
What is the value of cosβ ?
Enter your answer, in simplest radical form

 

Hello Guest!

 

\(tan(\alpha)=\frac{2\sqrt{7}}{5}\\ \alpha=arctan(\frac{2\sqrt{7}}{5})\\ \beta=180^0-\alpha=180^0-arctan(\frac{2\sqrt{7}}{5})=180^0-46.622^0\\ \beta=133.378^0 \)

\(cos(\beta)=-0.6868\)

laugh  !

 Jan 23, 2020
 #2
avatar+8956 
+1

A slightly different solution:

 

\((-5):(2\sqrt{7})=cos(\beta):sin(\beta) )\)

                                            \(sin(\beta)=\sqrt{1-cos^2(\beta)}\)        

\((-5)/(2\sqrt{7})=cos(\beta)/\sqrt{1-cos^2(\beta)}\\ \frac{25}{28}=\frac{cos^2(\beta)}{1-cos^2(\beta)}\)

\(28cos^2(\beta)=25-25cos^2(\beta)\)

\(53cos^2(\beta)=25\\ cos^2(\beta)=\frac{25}{53}\)

\(cos(\beta)=\pm\sqrt{\frac{25}{53}}=\pm \frac{5}{\sqrt{53}}\)

\(cos(\beta\ [-5;2\sqrt{7}])=- 0.686802819743\)      

\([\beta =133^022'39'']\)

 

laugh  !

 Jan 23, 2020
edited by asinus  Jan 23, 2020

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