Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.8 - Applications of Trigonometric Functions - Exercise Set - Page 637: 10


$ c=23.32;A=41.0^{\circ}; B=49.0^{\circ}$

Work Step by Step

The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ $\tan A=\dfrac{15.3}{17.6}=0.8693 \implies A=41.0^{\circ}$ So, $ B=90^{\circ}- 41.0^{\circ}=49.0^{\circ}$ and $ c=\dfrac{17.6}{\cos 41.0^{\circ}}=23.32$ Thus, $ c=23.32;A=41.0^{\circ}; B=49.0^{\circ}$
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