Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 11 - Section 11.3 - The Tangent Ratio and Other Ratios - Exercises - Page 511: 3


$\tan\alpha=\frac{\sqrt5}{2}$ $\tan\beta=\frac{2}{\sqrt5}$

Work Step by Step

Let $x$ be the side of the triangle with an unknown length. Using Pythagorean Theorem to find $x$. $x=\sqrt{6^{2}-4^{2}}$ $x=\sqrt{20}$ $x=2\sqrt5$ $\tan\alpha=\frac{opposite}{hypotenuse}=\frac{x}{4}=\frac{2\sqrt5}{4}=\frac{\sqrt5}{2}$ $\tan\beta=\frac{opposite}{hypotenuse}=\frac{4}{x}=\frac{4}{2\sqrt5}=\frac{2}{\sqrt5}$
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