Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Section 5.5 - Special Right Triangles - Exercises - Page 247: 12

Answer

$DE=6\sqrt 3 \approx 10.39$ $DF=6\sqrt 3\times \sqrt 3 = 18$

Work Step by Step

By the given Figure $ m\angle E = 2 .m\angle F$ So $ m\angle E = 60^{\circ} $ and $ m\angle F =30^{\circ}$ $EF=12\sqrt 3$ given Now by Theorem 5.5.2 / (30-60-90 Theorem) $DE=6\sqrt 3 \approx 10.39$ $DF=6\sqrt 3\times \sqrt 3 = 18$
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