SAT Question 244ff6c4: Circles answer and explanation with Desmos solution

Choice A is correct. A trigonometric ratio can be found using the unit circle, that is, a circle with radius $1$ unit. If a central angle of a unit circle in the xy-plane centered at the origin has its starting side on the positive x-axis and its terminal side intersects the circle at a point $\left(x,y\right)$, then the value of the tangent of the central angle is equal to the y-coordinate divided by the x-coordinate. There are $2\pi$ radians in a circle. Dividing $\frac{92\pi }{3}$ by $2\pi$ yields $\frac{92}{6}$, which is equivalent to $15+\frac{2}{3}$. It follows that on the unit circle centered at the origin in the xy-plane, the angle $\frac{92\pi }{3}$ is the result of $15$ revolutions from its starting side on the positive x-axis followed by a rotation through $\frac{2\pi }{3}$ radians. Therefore, the angles $\frac{92\pi }{3}$ and $\frac{2\pi }{3}$ are coterminal angles and $\tan \left(\frac{92\pi }{3}\right)$ is equal to $\tan \left(\frac{2\pi }{3}\right)$. Since $\frac{2\pi }{3}$ is greater than $\frac{\pi }{2}$ and less than $\pi$, it follows that the terminal side of the angle is in quadrant II and forms an angle of $\frac{\pi }{3}$, or $60^{\circ }$, with the negative x-axis. Therefore, the terminal side of the angle intersects the unit circle at the point $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$. It follows that the value of $\tan \left(\frac{2\pi }{3}\right)$ is $\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$, which is equivalent to $-\sqrt{3}$. Therefore, the value of $\tan \left(\frac{92\pi }{3}\right)$ is $-\sqrt{3}$.