Algebra 2 (1st Edition)

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

Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Problem Solving - Page 872: 39c


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Work Step by Step

We know that $d=\dfrac{v^2}{32} \sin (2 \theta)$ $\dfrac{v^2}{9} [\sin 2 (45-k)^{\circ}]= \dfrac{v^2}{9} [\sin 2 (45+k)^{\circ}]$ or, $\dfrac{v^2}{9} \sin (90-2k)^{\circ}= \dfrac{v^2}{9} \sin (90+2k)^{\circ}$ or, $\dfrac{v^2}{9} \cos (2k)^{\circ}= \dfrac{v^2}{9} \cos (2k)^{\circ}$ Hence, the horizontal distance when $\theta= (45-k)^{\circ} $ with distance when $\theta= (45+k)^{\circ} $ is the same.
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