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} . $\newline$ $P=\left(\frac{\sqrt{2}}{5}, \frac{\sqrt{23}}{5}\right)$ $\newline$ Answer:

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Algebra 2

Find trigonometric ratios using the unit circle

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Q. 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{2}}{5}, \frac{\sqrt{23}}{5}\right)

\newline

Answer:

Identify Coordinates: Identify the coordinates of the point on the unit circle. $\newline$ The given point P has coordinates $(x, y) = (\left(\frac{\sqrt{2}}{5}\right), \left(\frac{\sqrt{23}}{5}\right))$ . On the unit circle, these coordinates correspond to $(\cos(\theta), \sin(\theta))$ .
Determine Quadrant: Determine the quadrant in which the angle $\theta$ lies. $\newline$ Since both $x$ and $y$ coordinates are positive, the point lies in the first quadrant, where angle $\theta$ ranges from $0$ to $90$ degrees.
Calculate Angle: Calculate the angle $\theta$ using the inverse trigonometric function. $\newline$ To find $\theta$ , we can use the inverse tangent function because it gives us the angle whose tangent is the ratio of the y-coordinate to the x-coordinate. Thus, $\theta = \text{arctan}\left(\frac{\sqrt{23}}{5} / \frac{\sqrt{2}}{5}\right)$ .
Perform Calculation: Perform the calculation for $\theta$ . $\theta = \arctan\left(\frac{\sqrt{23}}{5} / \frac{\sqrt{2}}{5}\right) = \arctan\left(\frac{\sqrt{23}}{\sqrt{2}}\right) = \arctan\left(\sqrt{\frac{23}{2}}\right)$ . Now, we use a calculator to find the value of $\theta$ to the nearest tenth of a degree.
Convert to Degrees: Convert the angle from radians to degrees (if necessary) and round to the nearest tenth. $\newline$ Assuming the calculator is set to degree mode, we find that $\theta \approx \arctan(\sqrt{23/2}) \approx 67.4$ degrees (rounded to the nearest tenth).