Let (-sqrt (11),5) be a point on the terminal side of Theta Find the exact values of sinTheta ,cscTheta , and cotTheta sinTheta = cscTheta = cotTheta =

To find the exact values of $\sin\Theta$, $\csc\Theta$, and $\cot\Theta$ for the point $(-\sqrt{11}, 5)$ on the terminal side of $\Theta$, we need to use the definitions of trigonometric functions in terms of the coordinates of a point on the terminal side of an angle.<br /><br />Step 1: Find the radius (r) using the Pythagorean theorem:<br />$r = \sqrt{x^2 + y^2}$<br />$r = \sqrt{(-\sqrt{11})^2 + 5^2}$<br />$r = \sqrt{11 + 25}$<br />$r = \sqrt{36}$<br />$r = 6$<br /><br />Step 2: Find $\sin\Theta$:<br />$\sin\Theta = \frac{y}{r}$<br />$\sin\Theta = \frac{5}{6}$<br /><br />Step 3: Find $\csc\Theta$:<br />$\csc\Theta = \frac{1}{\sin\Theta}$<br />$\csc\Theta = \frac{1}{\frac{5}{6}}$<br />$\csc\Theta = \frac{6}{5}$<br /><br />Step 4: Find $\cot\Theta$:<br />$\cot\Theta = \frac{x}{y}$<br />$\cot\Theta = \frac{-\sqrt{11}}{5}$<br /><br />Therefore, the exact values are:<br />$\sin\Theta = \frac{5}{6}$<br />$\csc\Theta = \frac{6}{5}$<br />$\cot\Theta = \frac{-\sqrt{11}}{5}$