Answer:
[tex]\cos{\theta} = \frac{\sqrt{5}}{3}[/tex]
Step-by-step explanation:
In the fourth quadrant, the sine is negative and the cosine is positive.
We also have that, for each angle [tex]\theta[/tex],
[tex]\sqrt{\sin^{2}{\theta} + \cos^{2}{\theta}} = 1[/tex]
So
[tex]\sqrt{-(\frac{2}{3})^{2} + \cos^{2}{\theta}}^{2} = 1^{2}[/tex]
[tex]\frac{4/9} + \cos^{2}{\theta} = 1[/tex]
[tex]\cos^{2}{\theta} = \frac{5}{9}[/tex]
[tex]\cos{\theta} = \frac{\sqrt{5}}{3}[/tex]
