nvert the rectangular coordinates \( (-3,3 \sqrt{3}) \) of a point to its polar coordinates \( (r, \theta) \mathrm{w} \) 0 and \( 0 \leq \theta<2 \pi \). oose the correct option: a. \( (r, \theta)=\left(6, \frac{5 \pi}{3}\right) \) b. \( (r, \theta)=\left(6, \frac{2 \pi}{3}\right) \times \) c. \( (r, \theta)=\left(6, \frac{5 \pi}{6}\right) \) d. \( (r, \theta)=\left(6, \frac{7 \pi}{6}\right) \) e. None of the options

To convert rectangular coordinates \((-3, 3\sqrt{3})\) to polar coordinates \((r, \theta)\): First, we calculate \( r \): \[ r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6. \] Next, we find \( \theta \) using the tangent function: \[ \tan(\theta) = \frac{y}{x} = \frac{3\sqrt{3}}{-3} = -\sqrt{3}. \] The reference angle whose tangent is \(\sqrt{3}\) is \(\frac{\pi}{3}\). Since the coordinates \((-3, 3\sqrt{3})\) are located in the second quadrant (where x is negative and y is positive), we adjust: \[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}. \] Thus, the polar coordinates are \((r, \theta) = (6, \frac{2\pi}{3})\). The correct option is: **b. \( (r, \theta) = \left(6, \frac{2\pi}{3}\right) \)**.