Precalculus (6th Edition) Blitzer

$$\sin \theta =\frac{5}{6}, \quad \cos \theta =\frac{\sqrt{11}}{6}, \\ \tan \theta = \frac{5\sqrt{11}}{11}, \quad \cot \theta = \frac{\sqrt{11}}{5}, \\ \sec \theta = \frac{6 \sqrt{11}}{11}, \quad \csc\theta = \frac{6}{5}$$
We should first find$AC$ by applying the Pythagorean Theorem:$$AC=\sqrt{(AB)^2-(BC)^2}=\sqrt{6^2-5^2}=\sqrt{11}$$So we can find the value of the trigonometric functions as follows.$$\sin \theta =\frac{BC}{AB}=\frac{5}{6}, \quad \cos \theta =\frac{AC}{AB}=\frac{\sqrt{11}}{6}, \\ \tan \theta =\frac{BC}{AC}=\frac{5}{\sqrt{11}}=\frac{5\sqrt{11}}{11}, \quad \cot \theta =\frac{AC}{BC}=\frac{\sqrt{11}}{5}, \\ \sec \theta =\frac{AB}{AC}=\frac{6}{\sqrt{11}}=\frac{6\sqrt{11}}{11}, \quad \csc \theta =\frac{AB}{BC}=\frac{6}{5}$$