Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 11 - Section 11.1 - The Sine Ratio and Applications - Exercises - Page 504: 17

Answer

$a\approx10.9$ ft $b\approx11.7$ ft

Work Step by Step

$\sin47^{\circ}=\frac{b}{16}$ $b=16\times\sin47^{\circ}$ $b\approx11.7$ Using Pythagorean Theorem to find $a$. $a=\sqrt {16^{2}-b^{2}}$ $a\approx\sqrt{16^{2}-11.7^{2}}$ $a\approx10.9$
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