Trigonometry (10th Edition)

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

Chapter 2 - Acute Angles and Right Triangles - Section 2.4 Solving Right Triangles - 2.4 Exercises - Page 73: 12

Answer

$b \approx 58.2 km$ $\angle B = 58^{\circ} 20'$ $c \approx 68.4 km$

Work Step by Step

1. $\angle B = 90^{\circ} -31^{\circ} 40' = 58^{\circ}20'$ 2. $\tan 31^{\circ} 40' = \frac{35.9}{b}$ $b= 35.9 \tan 31^{\circ} 40'\approx 58.2 km$ 3.$\sin 31^{\circ} 40' =\frac{35.9}{c}$ $c = 35.9 \sin 31^{\circ} 40' \approx 68.4km$
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