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: 11


$DE = 12$ $FE= 24$

Work Step by Step

By the given Figure $\angle E= 60^{\circ}$ So remaining $\angle F= 30^{\circ}$ $FD= 12 \sqrt 3$ given Now by Theorem 5.5.2 / (30-60-90 Theorem) $DE = 12$ $FE= 24$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.