Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Acute Angles and Right Triangles - Section 2.4 Solving Right Triangles - 2.4 Exercises - Page 78: 20

Answer

$\angle A \approx 18.28^{\circ}$ $\angle B \approx 71. 72^{\circ}$ $ b \approx 14.5 m$

Work Step by Step

1. $a = 4.8 m$; $c =15.3 m $ $\angle C = 90^{\circ}$ 2. $\sin A = \frac{a}{c}= \frac{4.80}{15.3}$ $\angle A = \sin^{-1} (0.0.3137254902) \approx 18.28^{\circ}$ 3. $\angle A+ \angle B = 90^{\circ}$ $ \angle B = 90^{\circ} -18.28^{\circ} = 71.72^{\circ}$ 4. Pythagoream theorem $15.3^2 = 4.80^2 + b^2$ $b = \sqrt {234.09 -23.04}$ $b \approx 14.5 m$ consider only positive root, as length always positive in trigonometry
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