Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 11 - Section 11.3 - The Tangent Ratio and Other Ratios - Exercises - Page 511: 9


$\sin\alpha=\frac{x}{\sqrt{x^{2}+1}}$ $\cos\alpha=\frac{1}{\sqrt{x^{2}+1}}$ $\tan\alpha=x$ $\csc\alpha=\frac{\sqrt{x^{2}+1}}{x}$ $\sec\alpha=\sqrt{x^{2}+1}$ $\cot\alpha=\frac{1}{x}$

Work Step by Step

Let $l$ be the length of the side with an unknown length. $l=\sqrt{(\sqrt{x^{2}+1})^{2}-x^{2}}$ $l=\sqrt{x^{2}+1-x^{2}}$ $l=\sqrt{1}$ $l=1$ $\sin\alpha=\frac{opposite}{hypotenuse}=\frac{x}{\sqrt{x^{2}+1}}$ $\cos\alpha=\frac{adjacent}{hypotenuse}=\frac{1}{\sqrt{x^{2}+1}}$ $\tan\alpha=\frac{opposite}{adjacent}=\frac{x}{1}=x$ $\csc\alpha=\frac{hypotenuse}{opposite}=\frac{\sqrt{x^{2}+1}}{x}$ $\sec\alpha=\frac{hypotenuse}{adjacent}=\frac{\sqrt{x^{2}+1}}{1}=\sqrt{x^{2}+1}$ $\cot\alpha=\frac{adjacent}{opposite}=\frac{1}{x}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.