Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 550: 109

Answer

The ratio of the length of the side opposite $\theta $ to the length of the hypotenuse is $\frac{\sqrt{2}}{2}$.

Work Step by Step

Consider the provided values, $a=1$ and $b=1$ Substitute $a=1$ and $b=1$ in the equation ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$. $\begin{align} & {{c}^{2}}={{1}^{2}}+{{1}^{2}} \\ & {{c}^{2}}=1+1 \\ & {{c}^{2}}=2 \end{align}$ Use the square root property. $c=\pm \sqrt{2}$ Sides of a triangle are always positive $c>0$. So, the value of $c$ is $\sqrt{2}$. The ratio of the length of the side opposite $\theta $ to the length of the hypotenuse is $\frac{a}{c}$. Substitute $a=1$ and $c=\sqrt{2}$ in the expression $\frac{a}{c}$. $\frac{a}{c}=\frac{1}{\sqrt{2}}$ Multiply the numerator and denominator by $\sqrt{2}$. $\begin{align} & \frac{a}{c}=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} \\ & =\frac{\sqrt{2}}{2} \end{align}$ Therefore, the ratio of the length of the side opposite $\theta $ to the length of the hypotenuse is $\frac{\sqrt{2}}{2}$.
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