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 871: 35


$ \approx 10.05$ ft more

Work Step by Step

Here, we have $\theta_1=45^{\circ}$ and $\theta_2=60^{\circ}$ We know that $d=\dfrac{v^2}{32} \sin (2 \theta)$ $d=\dfrac{49^2}{32} \sin 2 (45^{\circ}) \approx 75.03 $ft ...(1) and $d=\dfrac{49^2}{32} \sin 2 (60^{\circ}) \approx 64.98 $ft ...(2) Now, we get $75.03-64.98 \approx 10.05$ ft more
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