Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 1 - Introduction and Mathematical Concepts - Problems - Page 21: 18

Answer

$29.6^{\circ}, 51.2^{\circ}, 99.2^{\circ} $

Work Step by Step

$ \gamma = cos^{-1}(\frac{(150)^{2}+(190)^{2}-(95)^{2}}{(2)(150)(190)}) \approx 29.6^{\circ} $ $\alpha = cos^{-1}(\frac{(95)^{2}+(190)^{2}-(150)^{2}}{(2)(95)(190)}) \approx 51.2^{\circ} $ $\beta$ = 180 - (51.2) - (29.6) = 99.2$^{\circ}$
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