Elementary Geometry for College Students (7th Edition) Clone

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: 11

Answer

$DE = 12$ $FE= 24$

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) $DE = 12$ $FE= 24$
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