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: 69

Answer

(a) CH = $\sqrt 2$ (b) CF = $\sqrt 3$

Work Step by Step

(a) Each face of a cube is a square and a diagonal divides it into two 45°–45°–90° triangles. Therefore- In 45°–45°–90° triangle CDH Diagonal CH = $\sqrt 2 \times $ DH CH = $\sqrt 2 \times $ 1 = $\sqrt 2$ (b) CHF is a right triangle, right angled at H. Applying Pythagorean theorem- $CF^{2} $ = $CH^{2} $ + $HF^{2} $ $CF^{2} $ = $(\sqrt 2)^{2} $ + $1^{2} $ [HF is the side of cube] $CF^{2} $ = $2 + 1 $ = 3 Therefore- CF = $\sqrt 3$
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