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=(-(sqrt11)/(5),(sqrt14)/(5)) Answer:

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Identify Point Quadrant: To find the angle θ that corresponds to a point on the unit circle, we can use the inverse trigonometric functions. The point P has coordinates (-√11/5, √14/5). Since the x-coordinate is negative and the y-coordinate is positive, the point lies in the second quadrant.Calculate Reference Angle: We can use the inverse tangent function to find the reference angle α, which is the acute angle to the x-axis. However, since the tangent function only gives us angles in the first and fourth quadrants, we need to adjust our approach. We can use the y-coordinate and x-coordinate to find the tangent of the reference angle α: tan(α) = |y/x| = |√14/5 / -√11/5| = √14/√11.Find Reference Angle: Now we calculate the reference angle α using the inverse tangent function: α = arctan(√14/√11). We use a calculator to find this angle.Calculate Actual Angle: After calculating, we find that α ≈ arctan(√(14/11)) ≈ 51.3 degrees. This is the reference angle in the first quadrant.Subtract Reference Angle: Since the point is in the second quadrant, we find the actual angle θ by subtracting the reference angle from 180 degrees: θ = 180 - α.Subtracting the reference angle from 180 degrees, we get θ ≈ 180 - 51.3 ≈ 128.7 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} . $\newline$ $P=\left(-\frac{\sqrt{11}}{5}, \frac{\sqrt{14}}{5}\right)$ $\newline$ Answer:

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