Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises - Page 531: 21

Answer

$b=\sqrt {c^{2}-a^{2}}=\sqrt {2^{2}-1^{2}}=\sqrt {3}$ $\sin B=\frac{\sqrt {3}}{2}$ $\cos B=\frac{1}{2}$ $\tan B=\sqrt 3$ $\csc B=\frac{2\sqrt {3}}{3}$ $\sec B=2$ $\cot B=\frac{\sqrt {3}}{3}$

Work Step by Step

$b=\sqrt {c^{2}-a^{2}}=\sqrt {2^{2}-1^{2}}=\sqrt {3}$ $\sin B=\frac{\text{side opposite}}{\text{hypotenuse}}=\frac{b}{c}=\frac{\sqrt {3}}{2}$ $\cos B=\frac{\text{side adjacent}}{\text{hypotenuse}}=\frac{a}{c}=\frac{1}{2}$ $\tan B=\frac{\text{side opposite}}{\text{side adjacent}}=\frac{b}{a}=\frac{\sqrt {3}}{1}=\sqrt 3$ The cosecant, secant, and cotangent ratios are reciprocals of the sine, cosine, and tangent values, respectively, so we have $\csc B=\frac{2}{\sqrt {3}}=\frac{2\sqrt {3}}{3}$ $\sec B=\frac{2}{1}=2$ $\cot B=\frac{1}{\sqrt {3}}=\frac{\sqrt {3}}{3}$
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