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: 17


$x=8 \sqrt 3; y=16$

Work Step by Step

Consider the lengths of the sides of the triangle, $x$ and $y$. $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{x}{h}$ and $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{y}{h}$ Here, $\sin 60^{\circ}=\dfrac{y}{h} \implies\sqrt 3=\dfrac{x}{ {8}}$ This gives: $x = 8 \sqrt 3$ Now, $ \dfrac{2}{1}=\dfrac{y}{8}$ This gives: $y=8 \times 2 =16$ Hence, $x=8 \sqrt 3; y=16$
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