## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 107d

#### Answer

$~\theta = tan^{-1}(\frac{x}{x^2+2})$

#### Work Step by Step

Let $~A~$ be the angle from the horizontal to the top of the painting. Let $~B~$ be the angle from the horizontal to the bottom of the painting. Then $\theta = A-B$. We can find an expression for $tan~\theta$: $tan~\theta = tan(A-B)$ $tan~\theta = \frac{tan~A-tan~B}{1+tan~A~tan~B}$ $tan~\theta = \frac{\frac{2}{x}-\frac{1}{x}}{1+(\frac{2}{x})(\frac{1}{x})}$ $tan~\theta = \frac{\frac{1}{x}}{1+\frac{2}{x^2}}$ $tan~\theta = \frac{\frac{1}{x}}{\frac{x^2+2}{x^2}}$ $tan~\theta = \frac{x^2}{(x)(x^2+2)}$ $tan~\theta = \frac{x}{x^2+2}$ Therefore, $~\theta = tan^{-1}(\frac{x}{x^2+2})$

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