Answer
$DF = 6\sqrt 3 \approx 10.39$
$DE= 6$
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)
$DF = 6\sqrt 3 \approx 10.39$
$DE= 6$
Elementary Geometry for College Students (7th Edition)
by Alexander, Daniel C.; Koeberlein, Geralyn M.
Published by
Cengage
ISBN 10:
978-1-337-61408-5
ISBN 13:
978-1-33761-408-5
Chapter 5 - Section 5.5 - Special Right Triangles - Exercises - Page 259: 10
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