Question

The angle θ lies in Quadrant IV . sinθ=−2/3 What is cosθ?

Answer 1

Consider this option:
1) if the angle belongs to IV quadrant, it means cos is in [0;1];
2)  $cos \alpha= \sqrt{1-sin^2 \alpha}$
using this formula cosθ=√(5/9) ⇒ cosθ=√5/3

$Answer: \frac{ \sqrt{5}}{3}$

Answer 2

Answer:

$\cos{\theta} = \frac{\sqrt{5}}{3}$

Step-by-step explanation:

In the fourth quadrant, the sine is negative and the cosine is positive.

We also have that, for each angle $\theta$,

$\sqrt{\sin^{2}{\theta} + \cos^{2}{\theta}} = 1$

So

$\sqrt{-(\frac{2}{3})^{2} + \cos^{2}{\theta}}^{2} = 1^{2}$

$\frac{4/9} + \cos^{2}{\theta} = 1$

$\cos^{2}{\theta} = \frac{5}{9}$

$\cos{\theta} = \frac{\sqrt{5}}{3}$