Precalculus (6th Edition)

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

Chapter 5 - Trigonometric Functions - 5.4 Solutions and Applications off Right Triangles - 5.4 Exercises - Page 547: 23

Answer

$B=62.0^{\circ}$ $a=8.17\,ft$ $b=15.4\,ft$

Work Step by Step

The sum of angles in a triangle is $180^{\circ}$. $\implies$ Angle $B=180^{\circ}-(90^{\circ}+28.0^{\circ})=62.0^{\circ}$ $\sin A=\sin 28.0^{\circ}=\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{a}{17.4\,ft}$ $\implies a=\sin 28.0^{\circ}\times17.4\,ft=8.17\,ft$ $\cos A=\cos 28.0^{\circ}=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{b}{17.4\,ft}$ $\implies b=\cos 28.0^{\circ}\times17.4\,ft=15.4\,ft$
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