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The point $\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)$ is on the unit circle. What is the Put the matsere, in degrees, of the angle passing through this $360^{\circ}$ , $15$ point?

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Find trigonometric ratios using the unit circle

Full solution

Q. The point

\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)

is on the unit circle. What is the Put the matsere, in degrees, of the angle passing through this

360^{\circ}

,

15

point?

Identify coordinates: Identify the coordinates of the point on the unit circle. The given point is $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$ , which corresponds to $(\cos(\theta), \sin(\theta))$ on the unit circle.
Recognize trigonometric values: Recognize that the $x$ -coordinate represents $\cos(\theta)$ and the $y$ -coordinate represents $\sin(\theta)$ . For the given point, $\cos(\theta) = \frac{\sqrt{3}}{2}$ and $\sin(\theta) = -\frac{1}{2}$ .
Determine quadrant: Determine the quadrant in which the angle $\theta$ lies. Since $\cos(\theta)$ is positive and $\sin(\theta)$ is negative, the angle $\theta$ must be in the fourth quadrant.
Find reference angle: Find the reference angle $\theta'$ in the first quadrant that has the same sine and cosine values, ignoring signs. The reference angle corresponding to $\cos(\theta') = \frac{\sqrt{3}}{2}$ and $\sin(\theta') = \frac{1}{2}$ is $30$ degrees or $\frac{\pi}{6}$ radians.
Calculate actual angle: Calculate the actual angle $\theta$ in degrees, knowing that it is in the fourth quadrant. In the fourth quadrant, the angle $\theta$ is $360$ degrees minus the reference angle $\theta'$ . Therefore, $\theta = 360$ degrees $- 30$ degrees.
Perform subtraction: Perform the subtraction to find the measure of angle $\theta$ . $\theta = 360^\circ - 30^\circ = 330^\circ$ .

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