Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 11 - Section 11.3 - The Tangent Ratio and Other Ratios - Exercises - Page 518: 26

Answer

$α \approx 35^{\circ}$ $β \approx 55^{\circ}$

Work Step by Step

1. Find $α$ $tan = \frac{opposite}{adjacent}$ $tan(α) = \frac{\sqrt {3}}{\sqrt {6}}$ by GDC / calculator $α = 35.26438...^{\circ}$ $α \approx 35^{\circ}$ 2. Find $β$ $tan = \frac{opposite}{adjacent}$ $tan(β) = \frac{\sqrt {6}}{\sqrt {3}}$ by GDC / calculator $β = 54.7356...^{\circ}$ $β \approx 55^{\circ}$
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