Given the following point on the unit circle, find the angle, to the nearest tenth of a degree (if necessary), of the terminal side through that point, 0^(@)

Question

Given the following point on the unit circle, find the angle, to the nearest tenth of a degree (if necessary), of the terminal side through that point,  0^(@) <= theta < 360^(@).  P=((sqrt3)/(3),-(sqrt6)/(3)) Answer:

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Identify Point P: Identify the coordinates of point P on the unit circle. Point P has coordinates P = ((sqrt(3)/3), -(sqrt(6)/3)).Recognize Coordinates Correspondence: Recognize that the coordinates of point P correspond to the cosine and sine of the angle theta, respectively. Cos(theta) = (sqrt(3)/3) and Sin(theta) = -(sqrt(6)/3).Calculate Reference Angle: Calculate the reference angle theta' using the arccosine of the absolute value of the x-coordinate. Theta' = arccos(|(sqrt(3)/3)|).Perform Reference Angle Calculation: Perform the calculation for the reference angle. Theta' = arccos((sqrt(3)/3)). Using a calculator, we find that Theta' ≈ 54.7 degrees.Determine Point Quadrant: Determine the quadrant in which point P lies. Since the x-coordinate is positive and the y-coordinate is negative, point P lies in the fourth quadrant.Find Actual Angle: Find the actual angle theta by subtracting the reference angle from 360 degrees, as the point is in the fourth quadrant. Theta = 360 - Theta'.Perform Angle Calculation: Perform the calculation to find the angle theta. Theta = 360 - 54.7. Theta ≈ 305.3 degrees.

Given the following point on the unit circle, find the angle, to the nearest tenth of a degree (if necessary), of the terminal side through that point, 0^{\circ} \leq \theta<360^{\circ} . $\newline$ $P=\left(\frac{\sqrt{3}}{3},-\frac{\sqrt{6}}{3}\right)$ $\newline$ Answer:

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