## Trigonometry 7th Edition

Published by Cengage Learning

# Chapter 1 - Section 1.1 - Angles, Degrees, and Special Triangles - 1.1 Problem Set - Page 13: 67

#### Answer

a = $2 \sqrt 3$ b = $\sqrt 3$ $d = \frac{3}{\sqrt 2}$

#### Work Step by Step

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

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