Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.1 - Angles, Degrees, and Special Triangles - 1.1 Problem Set: 44

Answer

BD = $\sqrt 89 \approx 9.43$

Work Step by Step

In given figure 23, we can see two right triangles BCD and BCA, both right angled at C. Given- BC = 5, AB = 13 , AD = 4 Applying Pythagoras theorem in right triangle BCA- $ AB^{2} $ = $ BC^{2} $ + $ AC^{2} $ $ AC^{2} $ = $ AB^{2} $ - $ BC^{2} $ $ AC^{2} $ = $ 13^{2} $ - $ 5^{2} $ $ AC^{2} $ = 169 - 25 = 144 $ AC = \sqrt (144)$ = 12 Now we can calculate DC as- DC = AC - AD = 12 - 4 = 8 Now applying Pythagoras theorem in right triangle BCD- $ BD^{2} $ = $ BC^{2} $ + $ DC^{2} $ $ BD^{2} $ = $ 5^{2} $ + $ 8^{2} $ $ BD^{2} $ = 25 + 64 = 89 Therefore BD = $\sqrt 89 \approx 9.43$
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