Given the following point on the unit circle, find the angle,to the nearest tenth necessary), of the terminal side through that point, 0^circ leqslant Theta lt 360^circ P=((sqrt (19))/(5),(sqrt (6))/(5)) Answer Attempt Lout of a

To find the angle $\Theta$ corresponding to the point $P=(\frac{\sqrt{19}}{5}, \frac{\sqrt{6}}{5})$ on the unit circle, we can use the inverse tangent function.<br /><br />Step 1: Calculate the tangent of the angle $\Theta$.<br />The tangent of an angle in the unit circle is given by the ratio of the y-coordinate to the x-coordinate. So, we have:<br />$\tan(\Theta) = \frac{\frac{\sqrt{6}}{5}}{\frac{\sqrt{19}}{5}} = \frac{\sqrt{6}}{\sqrt{19}}$<br /><br />Step 2: Use the inverse tangent function to find the angle $\Theta$.<br />The inverse tangent function, also known as arctan or $\tan^{-1}$, gives the angle whose tangent is a given value. So, we have:<br />$\Theta = \tan^{-1}\left(\frac{\sqrt{6}}{\sqrt{19}}\right)$<br /><br />Step 3: Calculate the value of $\Theta$.<br />Using a calculator, we can find the value of $\Theta$:<br />$\Theta \approx \tan^{-1}\left(\frac{\sqrt{6}}{\sqrt{19}}\right) \approx 28.6^{\circ}$<br /><br />Therefore, the angle $\Theta$ corresponding to the point $P=(\frac{\sqrt{19}}{5}, \frac{\sqrt{6}}{5})$ on the unit circle is approximately $28.6^{\circ}$.