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: 10


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

Work Step by Step

By the given Figure $\angle F= 30^{\circ}$ So remaining $\angle F= 60^{\circ}$ FE= 12 given Now by Theorem 5.5.2 / (30-60-90 Theorem) $DF = 6\sqrt 3 \approx 10.39$ $DE= 6$
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