+0

# The terminal ray of angle

0
149
2

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

What is the value of cosβ ?

cosβ=

Jan 23, 2020

#1
+9113
+1

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

Hello Guest!

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

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

!

Jan 23, 2020
#2
+9113
+1

A slightly different solution:

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

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

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

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

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

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

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

$$[\beta =133^022'39'']$$

!

Jan 23, 2020
edited by asinus  Jan 23, 2020