# Chapter 11 - Section 11.3 - The Tangent Ratio and Other Ratios - Exercises: 6

$\sin\alpha=\frac{\sqrt{11}}{4}$ $\cos\alpha=\frac{\sqrt{5}}{4}$ $\tan\alpha=\frac{\sqrt{11}}{\sqrt{5}}$ $\csc\alpha=\frac{4}{\sqrt{11}}$ $\sec\alpha=\frac{4}{\sqrt{5}}$ $\cot\alpha=\frac{\sqrt{5}}{\sqrt{11}}$

#### Work Step by Step

Let $x$ be the length of the side with an unknown length. Using Pythagorean Theorem to find $x$. $x=\sqrt{4^{2}-\sqrt5^{2}}$ $x=\sqrt{16-5}$ $x=\sqrt{11}$ $\sin\alpha=\frac{opposite}{hypotenuse}=\frac{\sqrt{11}}{4}$ $\cos\alpha=\frac{adjacent}{hypotenuse}=\frac{\sqrt{5}}{4}$ $\tan\alpha=\frac{opposite}{adjacent}=\frac{\sqrt{11}}{\sqrt{5}}$ $\csc\alpha=\frac{hypotenuse}{opposite}=\frac{4}{\sqrt{11}}$ $\sec\alpha=\frac{hypotenuse}{adjacent}=\frac{4}{\sqrt{5}}$ $\cot\alpha=\frac{adjacent}{opposite}=\frac{\sqrt{5}}{\sqrt{11}}$

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