Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set: 11

Answer

$\sqrt2$

Work Step by Step

RECALL: (1) $\sin{\theta} = \dfrac{\text{opposite side}}{\text{hypotenuse}}$ (2) $\tan{\theta} = \dfrac{\text{opposite side}}{\text{adjacent side}}$ (3) $\cos{\theta} = \dfrac{\text{adjacent side}}{\text{hypotenuse}}$ (4) $\sec{\theta} = \dfrac{\text{hypotenuse}}{\text{adjacent}}$ (5) $\csc{\theta} = \dfrac{\text{hypotenuse}}{\text{opposite side}}$ Use formula (4) above. Use the 45-degree angle of the triangle on the left to obtain: $\sec{45^o} = \dfrac{\sqrt2}{1}=\sqrt2$
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