Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.1 Use Trigonometry with Right Triangles - 13.1 Exercises - Skill Practice - Page 856: 3


$\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{24}{26}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{10}{26}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{24}{10}$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{26}{24}$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{26}{10}$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{10}{24}$

Work Step by Step

Use the Pythagorean Theorem to find the Hypotenuse. $Hypotenuse=\sqrt {24^2+10^2}=\sqrt {576+100}=26$ $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{24}{26}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{10}{26}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{24}{10}$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{26}{24}$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{26}{10}$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{10}{24}$
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.