Chapter 8 - Triangle Trigonometry and Polar Coordinates - 8.3 Polar Coordinates - Exercises and Problems for Section 8.3 - Exercises and Problems - Page 347: 22

$$\left(-\sqrt{3},\:1\right)$$

Converting the polar coordinates to Cartesian coordinates, we find: $$x=2\cos \left(\left(\frac{5\pi }{6}\right)\right) =-\sqrt3\\ y=2\sin \left(\left(\frac{5\pi }{6}\right)\right) =1 \\ \left(-\sqrt{3},\:1\right)$$