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 - 13.4 Exercises - Problem Solving - Page 879: 39


$ \theta =\tan^{-1} (\dfrac{44t}{100})$ or, $\tan^{-1} (\dfrac{11t}{25})$

Work Step by Step

Remember: $ 30 mph= 44 ft/sec$ Now, $\tan \theta= 44t \times \dfrac{1}{100}=\dfrac{44t}{100}$ Re-write as: $\theta = \tan^{-1} (\dfrac{44t}{100})$ This implies that $ \theta =\tan^{-1} (\dfrac{44t}{100})$ or, $\tan^{-1} (\dfrac{11t}{25})$
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