Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 11 - Review Exercises - Page 526: 33

Answer

$\csc\theta=\frac{29}{20}$ $\sin\theta=\frac{20}{29}$

Work Step by Step

$\cot\theta=\frac{21}{20}$ $1+\cot^2\theta=\csc^2\theta$ $1+\frac{441}{400}=\csc^2\theta$ $\csc^2\theta=\frac{841}{400}$ $\csc\theta=\frac{29}{20}$ $\sin\theta=\frac{1}{\csc\theta}$ $\sin\theta=\frac{20}{29}$
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