1. **State the problem:** We are given a right triangle with vertices J, K, and L, right angle at J, and side lengths JK = 39, JL = 52, KL = 65. We know \(\cos K = \frac{3}{5}\) and want to find \(\sin L\) and express the trigonometric ratios in simplest fractional form. 2. **Identify sides relative to angles:** - Angle K is at vertex K, so the side opposite K is JL = 52. - Angle L is at vertex L, so the side opposite L is JK = 39. - The hypotenuse is KL = 65. 3. **Check the given cosine ratio for angle K:** \[ \cos K = \frac{\text{adjacent to } K}{\text{hypotenuse}} = \frac{JK}{KL} = \frac{39}{65} \] Simplify: \[ \frac{39}{65} = \frac{\cancel{13}3}{\cancel{13}5} = \frac{3}{5} \] This matches the given \(\cos K = \frac{3}{5}\). 4. **Find \(\sin K\):** Using Pythagorean identity: \[ \sin^2 K + \cos^2 K = 1 \] \[ \sin^2 K = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \] \[ \sin K = \frac{4}{5} \] 5. **Find \(\sin L\):** Since angle L is complementary to angle K in a right triangle, \(\sin L = \cos K = \frac{3}{5}\). 6. **Express all trig ratios for angles K and L:** - \(\cos K = \frac{3}{5}\) - \(\sin K = \frac{4}{5}\) - \(\sin L = \frac{3}{5}\) - \(\cos L = \sin K = \frac{4}{5}\) **Final answer:** \[ \cos K = \frac{3}{5}, \quad \sin L = \frac{3}{5} \]