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: 70

Answer

(a) GD = $5\sqrt 2$ (b) GB = $5\sqrt 3$

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 GCD Diagonal GD = $\sqrt 2 \times $ CD GD = $\sqrt 2 \times $ 5 ( CD is the edge of cube given as 5) GD = $5\sqrt 2$ (b) GDB is a right triangle, right angled at D. Applying Pythagorean theorem- $GB^{2} $ = $GD^{2} $ + $DB^{2} $ $GB^{2} $ = $(5\sqrt 2)^{2} $ + $5^{2} $ [DB is the side of cube given equal to 5] $GB^{2} $ = $50 + 25 $ = 75 Therefore- GB = $\sqrt 75$ = $\sqrt (25\times3)$ GB = $\sqrt 25 \times \sqrt 3$ GB = $5\sqrt 3$
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