Precalculus (6th Edition)

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

Chapter 8 - Application of Trigonometry - Chapter 8 Test Prep - Review Exercises - Page 838: 17

Answer

$$B = 17.3^\circ ,\,\,\,C = 137.5^\circ ,\,\,\,\,c = 11{\text{yd}}$$

Work Step by Step

$$\eqalign{ & A = {\text{25}}.{\text{2}}^\circ ,a = {\text{6}}.{\text{92 yd}},b = {\text{4}}.{\text{82 yd}} \cr & {\text{Calculate the angle }}B{\text{ using the law of sines}} \cr & \frac{{\sin B}}{b} = \frac{{\sin A}}{a} \cr & \sin B = \frac{{b\sin A}}{a} \cr & \sin B = \frac{{{\text{4}}.{\text{82}}\sin \left( {{\text{25}}.{\text{2}}^\circ } \right)}}{{6.92}} \cr & {\text{Use a calculator}} \cr & \sin B \approx 0.2965688129 \cr & {\text{Use the inverse sine function}} \cr & B \approx 17.3^\circ \cr & \cr & {\text{Calculate the angle }}C \cr & C = 180^\circ - A - B \cr & C = 180^\circ - {\text{25}}.{\text{2}}^\circ - 17.3^\circ \cr & C = 137.5^\circ \cr & \cr & {\text{Calculate the side }}c{\text{ using the law of sines}} \cr & \frac{c}{{\sin C}} = \frac{a}{{\sin A}} \cr & c = \frac{{a\sin C}}{{\sin A}} \cr & c = \frac{{{\text{6}}.{\text{92}}\sin \left( {137.5^\circ } \right)}}{{\sin \left( {{\text{25}}.{\text{2}}^\circ } \right)}} \cr & c \approx 11{\text{yd}} \cr & \cr & {\text{Answer}} \cr & B = 17.3^\circ ,\,\,\,C = 137.5^\circ ,\,\,\,\,c = 11{\text{yd}} \cr} $$
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