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: 43

Answer

AB = $\sqrt 41 \approx 6.40$

Work Step by Step

In given figure 23, we can see two right triangles BCD and BCA, both right angled at C. Given- BC = 4, BD = 5 , AD = 2 In right triangle BCD, applying Pythagoras theorem- $ BD^{2} $ = $ BC^{2} $ + $ DC^{2} $ Therefore- $ DC^{2} $ = $ BD^{2} $- $ BC^{2} $ $ DC^{2} $ = $ 5^{2} $- $ 4^{2} $ = 25-16=9 $ DC = \sqrt 9$ = 3 Now we can calculate AC as- AC = AD + DC = 2 + 3 = 5 Now applying Pythagoras theorem in right triangle BCD- $ AB^{2} $ = $ BC^{2} $ + $ AC^{2} $ $ AB^{2} $ = $ 4^{2} $ + $ 5^{2} $ $ AB^{2} $ = 16 + 25 = 41 Therefore AB = $\sqrt 41 \approx 6.40$
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