Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises: 19

Answer

$a=\sqrt{21}$ $\sin B=\dfrac{2}{5}$ $;$ $\cos B=\dfrac{\sqrt{21}}{5}$ $;$ $\tan B=\dfrac{2\sqrt{21}}{21}$ $\csc B=\dfrac{5}{2}$ $;$ $\sec B=\dfrac{5\sqrt{21}}{21}$ $;$ $\cot B=\dfrac{\sqrt{21}}{2}$

Work Step by Step

$b=2$ $,$ $c=5$ Since $\angle C$ is the right angle, $c$ is the hypotenuse of this triangle and $b$ is a cathetus. Use the Pythagorean Theorem to obtain the unknown cathetus $a$: $a=\sqrt{c^{2}-b^{2}}$ $a=\sqrt{5^{2}-2^{2}}$ $a=\sqrt{25-4}$ $a=\sqrt{21}$ Proceed to find the six trigonometric functions for angle $B$: $\sin B=\dfrac{opposite}{hypotenuse}=\dfrac{2}{5}$ $\cos B=\dfrac{adjacent}{hypotenuse}=\dfrac{\sqrt{21}}{5}$ $\tan B=\dfrac{opposite}{adjacent}=\dfrac{2}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}$ $\csc B=\dfrac{hypotenuse}{opposite}=\dfrac{5}{2}$ $\sec B=\dfrac{hypotenuse}{adjacent}=\dfrac{5}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}$ $\cot B=\dfrac{adjacent}{opposite}=\dfrac{\sqrt{21}}{2}$
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