Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 282: 52a

Answer

$\theta = tan^{-1}(\frac{4}{x})-tan^{-1}(\frac{1}{x})$

Work Step by Step

Let $\alpha_1$ be the small angle between the horizontal line and the bottom of the painting. Let $\alpha_2$ be the large angle between the horizontal line and the top of the painting. $tan~\alpha_1 = \frac{1}{x}$ $\alpha_1 = tan^{-1}(\frac{1}{x})$ $tan~\alpha_2 = \frac{3+1}{x}$ $\alpha_2 = tan^{-1}(\frac{4}{x})$ We can write an expression for $\theta$: $\theta = \alpha_2-\alpha_1$ $\theta = tan^{-1}(\frac{4}{x})-tan^{-1}(\frac{1}{x})$
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