1. The problem asks for the trigonometric ratios \(\sin M\), \(\cos M\), and \(\tan M\) in the right triangle KLM with right angle at K. 2. Recall the definitions of the trigonometric ratios for angle \(M\): - \(\sin M = \frac{\text{opposite side}}{\text{hypotenuse}}\) - \(\cos M = \frac{\text{adjacent side}}{\text{hypotenuse}}\) - \(\tan M = \frac{\text{opposite side}}{\text{adjacent side}}\) 3. Identify the sides relative to angle \(M\): - Opposite side to \(M\) is \(KL = 2\sqrt{7}\) - Adjacent side to \(M\) is \(KM = 6\) - Hypotenuse is \(LM = 8\) 4. Calculate each ratio: \[\sin M = \frac{KL}{LM} = \frac{2\sqrt{7}}{8} = \frac{\cancel{2}\sqrt{7}}{\cancel{8}4} = \frac{\sqrt{7}}{4}\] \[\cos M = \frac{KM}{LM} = \frac{6}{8} = \frac{\cancel{6}}{\cancel{8}} \frac{3}{4}\] \[\tan M = \frac{KL}{KM} = \frac{2\sqrt{7}}{6} = \frac{\cancel{2}\sqrt{7}}{\cancel{6}3} = \frac{\sqrt{7}}{3}\] 5. Final answers: - \(\sin M = \frac{\sqrt{7}}{4}\) - \(\cos M = \frac{3}{4}\) - \(\tan M = \frac{\sqrt{7}}{3}\)