## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 13, Trigonometric Ratios and Functions - 13.1 Use Trigonometry with Right Triangles - 13.1 Exercises - Skill Practice - Page 857: 29b

#### Answer

$2n \sin (\dfrac{180}{n})^{\circ}$

#### Work Step by Step

Here, $\theta=(\dfrac{1}{n}) \times 360^{\circ}=(\dfrac{360}{n})^{\circ}$ $\sin \theta/2=\dfrac{x}{1}$ This gives: $\sin \dfrac{(\dfrac{360}{n})^{\circ}}{2}=x$ $\implies x=\sin (\dfrac{180}{n})^{\circ}$ Now, the perimeter of the n-sided polygon is: $P=n (2x)= 2n \sin (\dfrac{180}{n})^{\circ}$

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