Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 44


$\approx 42.22^o$

Work Step by Step

We know that the three angles spanning the line must add up to 180 degrees: $65^o (90^o-\beta)+(90^o-\alpha)=180^o $ Solving for $\alpha$, we get: $\alpha=65^o-\beta=65^o-tan^{-1}(\frac{21}{50})$ Which approximately equals: $\approx65^o - 22.78^o\approx42.22^o $
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