Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.2 - Right Triangle Trigonometry - 6.2 Exercises - Page 441: 8

Answer

$sin~θ=\frac{\sqrt 2}{2}$ $cos~θ=\frac{\sqrt 2}{2}$ $tan~θ=1$ $csc~θ=\sqrt 2$ $sec~θ=\sqrt 2$ $cot~θ=1$

Work Step by Step

First, let's evaluate the side opposite to $θ$: $(hyp)^2=(opp)^2+(adj)^2$ $(7\sqrt 2)^2=(opp)^2+7^2$ $49(2)-49=(opp)^2$ $49=(opp)^2$ $opp=7$ $sin~θ=\frac{opp}{hyp}=\frac{7}{7\sqrt 2}=\frac{1}{\sqrt 2}\frac{\sqrt 2}{\sqrt 2}=\frac{\sqrt 2}{2}$ $cos~θ=\frac{adj}{hyp}=\frac{7}{7\sqrt 2}=\frac{1}{\sqrt 2}\frac{\sqrt 2}{\sqrt 2}=\frac{\sqrt 2}{2}$ $tan~θ=\frac{opp}{adj}=\frac{7}{7}=1$ $csc~θ=\frac{hyp}{opp}=\frac{7\sqrt 2}{7}=\sqrt 2$ $sec~θ=\frac{hyp}{adj}=\frac{7\sqrt 2}{7}=\sqrt 2$ $cot~θ=\frac{adj}{opp}=\frac{7}{7}=1$
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