Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.1 Use Trigonometry with Right Triangles - 13.1 Exercises - Problem Solving - Page 857: 31


$63.39 cm$

Work Step by Step

Let us consider $x$ to be the distance. Since, $\sin \theta=\dfrac{Opposite}{Hypotensue}=\dfrac{x}{150}$ or, $\sin 25^{\circ}=\dfrac{x}{150}$ This gives: $ x =150 \sin 25^{\circ}$ or, $x \approx 63.39 cm$
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