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: 68

Answer

b = 2 c = 2$\sqrt 3$ and d = $\sqrt 6$

Work Step by Step

In given figure 28, 'a' is the longest side, 'b' is the shortest side and 'c' is the side opposite $60^{\circ}$ of a 30°–60°–90° triangle- The longest side of a 30°–60°–90° triangle is twice the shortest side and the side opposite the 60° angle is $\sqrt 3$ times the shortest side. Given that Longest side, a = 4 Therefore Shortest side, b = $\frac{longest side}{2}$ = $\frac{4}{2}$ = 2 Side opposite 60° i.e. c = $\sqrt 3\times $ shortest side = $\sqrt 3 \times$2 = 2$\sqrt 3$ Side opposite 60° i.e. c = 2$\sqrt 3$ Now 'c' is the hypotenuse and d is one of the shorter sides of a 45°–45°–90° triangle. Hence- c = $d\times\sqrt 2$ = $d\sqrt 2$ Therefore d =$ \frac{c}{\sqrt 2} $ = $ \frac{2\sqrt 3}{\sqrt 2} $ d = $ \frac{\sqrt 2\times\sqrt 2\times\sqrt 3}{\sqrt 2} $ ( writing 2 as $\sqrt 2\times\sqrt 2$) d= $\sqrt 2\times\sqrt 3$ = $\sqrt 6$
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