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: 16


$\dfrac{11 \sqrt 2}{12}$

Work Step by Step

We need to use the Pythagorean Theorem in order to solve $y$ such that $r^2=x^2+y^2 \implies y=\sqrt {r^2-x^2}$ $y=\sqrt{(11)^2-(7)^2}=\sqrt{72}=6\sqrt 2$ Since, $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{r}{y}$ Thus, $\csc \theta=\dfrac{11}{6\sqrt{2}}=\dfrac{11 \sqrt 2}{12}$
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.