Question

the diagram below shows an angle graphed in the xy-coordinate plane. segment rt is the initial side of the angle, and segment rm is the terminal side. segments rt and rm are radii of the unit circle centered at the origin r(0, 0). the x-coordinate of m is \\(\frac{\sqrt{3}}{2}\\). what is the measure of the angle to the nearest thousandth of a radian?

Explanation:

Step1: Recall Unit Circle Coordinates

On the unit circle, the coordinates of a point \( M(x, y) \) on the terminal side of an angle \( \theta \) (in radians) are given by \( x = \cos\theta \) and \( y = \sin\theta \). Here, the \( x \)-coordinate of \( M \) is \( \frac{\sqrt{3}}{2} \), so \( \cos\theta=\frac{\sqrt{3}}{2} \).

Step2: Find the Angle

We know that \( \cos\theta=\frac{\sqrt{3}}{2} \) and the angle is in the fourth quadrant (since the \( x \)-coordinate is positive and \( y \)-coordinate is negative, as seen from the diagram). The reference angle for which \( \cos\alpha=\frac{\sqrt{3}}{2} \) is \( \alpha = \frac{\pi}{6} \) (or \( 30^\circ \)). But since it's in the fourth quadrant, the angle \( \theta = 2\pi - \frac{\pi}{6}=\frac{11\pi}{6}\approx 5.7596 \)? Wait, no—wait, maybe I misread the diagram. Wait, the initial side is along the negative y - axis? Wait, no, the initial side RT: looking at the diagram, R is at (0,0), T is along the negative y - axis? Wait, no, the x - coordinate of M is \( \frac{\sqrt{3}}{2} \), and if the initial side is along the negative y - axis (RT), then we need to adjust. Wait, maybe the angle is measured from the negative y - axis. Wait, let's re - evaluate.

Wait, the unit circle: if the initial side is along the negative y - axis (downwards), then the angle \( \theta \) is measured from the negative y - axis to the terminal side RM. The x - coordinate of M is \( \frac{\sqrt{3}}{2} \), y - coordinate would be negative (since it's in the fourth quadrant relative to the standard position, but if initial side is negative y - axis, let's consider the angle \( \alpha \) from the negative y - axis. Let's use the coordinates: \( x=\cos\theta \), \( y = \sin\theta \), but if the initial side is along the negative y - axis, then the angle \( \theta \) (from negative y - axis) has \( x=\sin\theta \), \( y = -\cos\theta \)? Wait, maybe I made a mistake. Wait, let's use the standard position. Wait, the center R is at (0,0). The initial side RT: looking at the diagram, T is along the negative y - axis? No, the x - axis and y - axis: the vertical line is the y - axis, horizontal is x - axis. R is at (0,0), T is along the negative y - axis? Wait, no, the x - coordinate of M is \( \frac{\sqrt{3}}{2} \), so \( x=\cos\theta \), where \( \theta \) is the angle from the positive x - axis. Wait, but in the diagram, the initial side RT: maybe RT is along the negative y - axis. Wait, let's think again.

Wait, the problem says "the x - coordinate of M is \( \frac{\sqrt{3}}{2} \)". So \( x = \cos\theta=\frac{\sqrt{3}}{2} \), and the angle is in the fourth quadrant (since[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]

Answer:

Step1: Recall Unit Circle Coordinates

On the unit circle, the coordinates of a point \( M(x, y) \) on the terminal side of an angle \( \theta \) (in radians) are given by \( x = \cos\theta \) and \( y = \sin\theta \). Here, the \( x \)-coordinate of \( M \) is \( \frac{\sqrt{3}}{2} \), so \( \cos\theta=\frac{\sqrt{3}}{2} \).

Step2: Find the Angle

We know that \( \cos\theta=\frac{\sqrt{3}}{2} \) and the angle is in the fourth quadrant (since the \( x \)-coordinate is positive and \( y \)-coordinate is negative, as seen from the diagram). The reference angle for which \( \cos\alpha=\frac{\sqrt{3}}{2} \) is \( \alpha = \frac{\pi}{6} \) (or \( 30^\circ \)). But since it's in the fourth quadrant, the angle \( \theta = 2\pi - \frac{\pi}{6}=\frac{11\pi}{6}\approx 5.7596 \)? Wait, no—wait, maybe I misread the diagram. Wait, the initial side is along the negative y - axis? Wait, no, the initial side RT: looking at the diagram, R is at (0,0), T is along the negative y - axis? Wait, no, the x - coordinate of M is \( \frac{\sqrt{3}}{2} \), and if the initial side is along the negative y - axis (RT), then we need to adjust. Wait, maybe the angle is measured from the negative y - axis. Wait, let's re - evaluate.

Wait, the unit circle: if the initial side is along the negative y - axis (downwards), then the angle \( \theta \) is measured from the negative y - axis to the terminal side RM. The x - coordinate of M is \( \frac{\sqrt{3}}{2} \), y - coordinate would be negative (since it's in the fourth quadrant relative to the standard position, but if initial side is negative y - axis, let's consider the angle \( \alpha \) from the negative y - axis. Let's use the coordinates: \( x=\cos\theta \), \( y = \sin\theta \), but if the initial side is along the negative y - axis, then the angle \( \theta \) (from negative y - axis) has \( x=\sin\theta \), \( y = -\cos\theta \)? Wait, maybe I made a mistake. Wait, let's use the standard position. Wait, the center R is at (0,0). The initial side RT: looking at the diagram, T is along the negative y - axis? No, the x - axis and y - axis: the vertical line is the y - axis, horizontal is x - axis. R is at (0,0), T is along the negative y - axis? Wait, no, the x - coordinate of M is \( \frac{\sqrt{3}}{2} \), so \( x=\cos\theta \), where \( \theta \) is the angle from the positive x - axis. Wait, but in the diagram, the initial side RT: maybe RT is along the negative y - axis. Wait, let's think again.

Wait, the problem says "the x - coordinate of M is \( \frac{\sqrt{3}}{2} \)". So \( x = \cos\theta=\frac{\sqrt{3}}{2} \), and the angle is in the fourth quadrant (since[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]