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 248: 28

Answer

DC=$\frac{10\sqrt 3}{3}$$\approx$5.77 DB=$\frac{20\sqrt 3}{3}$$\approx$11.56

Work Step by Step

CB=AC$\sqrt 3$ CB=10$\sqrt 3$ DC$\sqrt 3$=AC DC=$\frac{10\sqrt 3}{3}$$\approx$5.77 DB=CB-DC DB=10$\sqrt 3$-$\frac{10\sqrt 3}{3}$ DB=$\frac{20\sqrt 3}{3}$$\approx$11.56
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