Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 11 - Section 11.3 - The Tangent Ratio and Other Ratios - Exercises - Page 511: 4

Answer

$tan(α) = \frac{\sqrt {3}}{2}$ $tan(β) = \frac{2}{\sqrt {3}}$

Work Step by Step

1. Find $AB$ using the Pythagorean Theorem $a^{2} + b^{2} = c^{2}$ $(AB)^{2} + (\sqrt {3})^{2} = (\sqrt {7})^{2}$ $(AB)^{2} + 3 = 7$ $(AB)^{2} = 4$ $AB = \sqrt {4}$ $AB = 2$ 2. Find $tan(α)$ (Recall that SOHCAHTOA, in this case we use TOA) $tan = \frac{opposite}{adjacent}$ $tan(α) = \frac{\sqrt {3}}{2}$ 2. Find $tan(β)$ $tan = \frac{opposite}{adjacent}$ $tan(β) = \frac{2}{\sqrt {3}}$
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