Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.1 Use Trigonometry with Right Triangles - 13.1 Exercises - Skill Practice - Page 856: 18


$x=y=\dfrac{\sqrt {42}}{2}$

Work Step by Step

Consider the lengths of the sides of the triangle, $x$ and $y$. $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{y}{h}$ and $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{x}{h}$ Here, $\sin 45^{\circ}=\dfrac{y}{h} \implies \dfrac{1}{\sqrt 2}=\dfrac{y}{\sqrt {21}}$ This gives: $y=\dfrac{\sqrt {42}}{2}$ Now, $\cos 45^{\circ}=\dfrac{x}{h} \implies \dfrac{1}{\sqrt 2}=\dfrac{x}{\sqrt {21}}$ This gives: $x=\dfrac{\sqrt {42}}{2}$ Hence, $x=y=\dfrac{\sqrt {42}}{2}$
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