# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 287: 51d

$\beta = arctan(\frac{(x+y)~tan~\alpha}{x})$

#### Work Step by Step

From the diagram: $tan~\alpha = \frac{x}{z}$ $z = \frac{x}{tan~\alpha}$ $tan~\beta = \frac{x+y}{z}$ $z = \frac{x+y}{tan~\beta}$ We can equate the two expressions for $z$: $\frac{x}{tan~\alpha} = \frac{x+y}{tan~\beta}$ $\frac{tan~\alpha}{tan~\beta} = \frac{x}{x+y}$ We can solve this equation for $\alpha$: $\frac{tan~\alpha}{tan~\beta} = \frac{x}{x+y}$ $tan~\beta = \frac{(x+y)~tan~\alpha}{x}$ $\beta = arctan(\frac{(x+y)~tan~\alpha}{x})$

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