Trigonometry (10th Edition)

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

Chapter 6 - Review Exercises - Page 286: 63a

Answer

Therefore, the value of $\theta$ is given by the function $f(x) = arctan(\frac{15}{x})-arctan(\frac{5}{x})$

Work Step by Step

Let $~A~$ be the angle between the wall and the right side of the board. $tan~A = \frac{15}{x}$ $A = arctan(\frac{15}{x})$ Let $~B~$ be the angle between the wall and the left side of the board. $tan~B = \frac{5}{x}$ $B = arctan(\frac{5}{x})$ We can find an expression for $\theta$: $\theta = A-B$ $\theta = arctan(\frac{15}{x})-arctan(\frac{5}{x})$ Therefore, the value of $\theta$ is given by the function $f(x) = arctan(\frac{15}{x})-arctan(\frac{5}{x})$
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