Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 8 - Right Triangles and Trigonometry - 8-2 Special Right Triangles - Practice and Problem-Solving Exercises - Page 504: 15


$x = 20$ $y = 20 \sqrt 3$

Work Step by Step

The diagram is that of a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle because one angle measures $30^{\circ}$ and another measures $90^{\circ}$, so the last angle must be $60^{\circ}$. In this type of right triangle, the hypotenuse is two times the shorter leg. Let's write an equation to solve for $x$, the length of the shorter leg: $40 = 2x$ Divide each side by $2$ to solve for $x$: $x = 20$ In this triangle, the longer leg is $\sqrt 3$ times the length of the shorter leg. Let's set up that equation to solve for $y$, the length of the longer leg: $y = 20 \sqrt 3$
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