Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 11 - Section 11.3 - The Tangent Ratio and Other Ratios - Exercises - Page 511: 5

Answer

$\sin\alpha=\frac{5}{13}$ $\cos\alpha=\frac{12}{13}$ $\tan\alpha=\frac{5}{12}$ $\csc\alpha=\frac{13}{5}$ $\sec\alpha=\frac{13}{12}$ $\cot\alpha=\frac{12}{5}$

Work Step by Step

Let $x$ be the length of the side with an unknown length. Using Pythagorean Theorem to find $x$. $x=\sqrt{13^{2}-5^{2}}$ $x=\sqrt{169-25}$ $x=\sqrt{144}$ $x=12$ $\sin\alpha=\frac{opposite}{hypotenuse}=\frac{5}{13}$ $\cos\alpha=\frac{adjacent}{hypotenuse}=\frac{12}{13}$ $\tan\alpha=\frac{opposite}{adjacent}=\frac{5}{12}$ $\csc\alpha=\frac{hypotenuse}{opposite}=\frac{13}{5}$ $\sec\alpha=\frac{hypotenuse}{adjacent}=\frac{13}{12}$ $\cot\alpha=\frac{adjacent}{opposite}=\frac{12}{5}$
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