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: 9


$DF = 5\sqrt 3 \approx 8.66$ $FE= 10$

Work Step by Step

By the given Figure $\angle D=90^{\circ}$ given $\angle E=60^{\circ}$ given So remaining$ \angle F=30^{\circ}$ now $DE= 5 $ given By Theorem 5.5.2/ (30-60-90 Theorem) $DF = 5\sqrt 3 \approx 8.66$ $FE= 10$
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