The point given below is on the terminal side of an angle \( e \). Find the exact value of eachy of the six trigonometric functions of \( \theta \). \( (-24,10) \) \( \sin \theta=\square \) (Simplify your answer. Type an exact answer, using radicals as noeded. Use integens or fractions for any numbers in the expression. Rationalize all denominators.)

To find the six trigonometric functions of the angle \( \theta \) whose terminal side passes through the point \( (-24, 10) \), we first need to determine the radius \( r \) using the coordinates of the point. The radius \( r \) is calculated using the formula: \[ r = \sqrt{x^2 + y^2} \] where \( x = -24 \) and \( y = 10 \). Calculating \( r \): \[ r = \sqrt{(-24)^2 + (10)^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \] Now we can find the six trigonometric functions: 1. **Sine**: \[ \sin \theta = \frac{y}{r} = \frac{10}{26} = \frac{5}{13} \] 2. **Cosine**: \[ \cos \theta = \frac{x}{r} = \frac{-24}{26} = -\frac{12}{13} \] 3. **Tangent**: \[ \tan \theta = \frac{y}{x} = \frac{10}{-24} = -\frac{5}{12} \] 4. **Cosecant** (the reciprocal of sine): \[ \csc \theta = \frac{r}{y} = \frac{26}{10} = \frac{13}{5} \] 5. **Secant** (the reciprocal of cosine): \[ \sec \theta = \frac{r}{x} = \frac{26}{-24} = -\frac{13}{12} \] 6. **Cotangent** (the reciprocal of tangent): \[ \cot \theta = \frac{x}{y} = \frac{-24}{10} = -\frac{12}{5} \] Now we can summarize the values of the six trigonometric functions: - \( \sin \theta = \frac{5}{13} \) - \( \cos \theta = -\frac{12}{13} \) - \( \tan \theta = -\frac{5}{12} \) - \( \csc \theta = \frac{13}{5} \) - \( \sec \theta = -\frac{13}{12} \) - \( \cot \theta = -\frac{12}{5} \) Thus, the exact values of the six trigonometric functions of \( \theta \) are: - \( \sin \theta = \frac{5}{13} \) - \( \cos \theta = -\frac{12}{13} \) - \( \tan \theta = -\frac{5}{12} \) - \( \csc \theta = \frac{13}{5} \) - \( \sec \theta = -\frac{13}{12} \) - \( \cot \theta = -\frac{12}{5} \)