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=
+14903 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!
\(tan(\alpha)=\frac{2\sqrt{7}}{5}\\ \alpha=arctan(\frac{2\sqrt{7}}{5})\\ \beta=180^0-\alpha=180^0-arctan(\frac{2\sqrt{7}}{5})=180^0-46.622^0\\ \beta=133.378^0 \)
\(cos(\beta)=-0.6868\)
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+14903 A slightly different solution:
\((-5):(2\sqrt{7})=cos(\beta):sin(\beta) )\)
\(sin(\beta)=\sqrt{1-cos^2(\beta)}\)
\((-5)/(2\sqrt{7})=cos(\beta)/\sqrt{1-cos^2(\beta)}\\ \frac{25}{28}=\frac{cos^2(\beta)}{1-cos^2(\beta)}\)
\(28cos^2(\beta)=25-25cos^2(\beta)\)
\(53cos^2(\beta)=25\\ cos^2(\beta)=\frac{25}{53}\)
\(cos(\beta)=\pm\sqrt{\frac{25}{53}}=\pm \frac{5}{\sqrt{53}}\)
\(cos(\beta\ [-5;2\sqrt{7}])=- 0.686802819743\)
\([\beta =133^022'39'']\)
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