Physics (10th Edition)

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

Chapter 1 - Introduction and Mathematical Concepts - Problems: 12


$0.528\; km$

Work Step by Step

Let $x$ = distance from base of arch $tan(20^{\circ}) = \frac{192}{x} $ $x = \frac{192}{tan(20^{\circ})} $ $x$ $ \approx 527.52\; m$ $= 527.52\; m \times \frac{1\; km}{1000\; m}$ $= 0.528\; km$
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