Geometry: Common Core (15th Edition)

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

Chapter 3 - Parallel and Perpendicular Lines - 3-6 Constructing Parallel and Perpendicular Lines - Practice and Problem-Solving Exercises: 43

Answer

3y = 120$^{\circ}$ (y - 15) = 25$^{\circ}$

Work Step by Step

All angles in a triangle must add up to 180∘. We can therefore add the three given angle measures together, set them equal to 180∘, and then solve for y. 35 + 3y + (y-15) = 180 35 - 35 + 3y + (y-15) = 180 - 35 3y + (y-15) = 145 (We can re-write (y-15) as y - 15.) 3y + y - 15 = 145 3y + y - 15 + 15 = 145 + 15 3y + y = 160 We can now combine like-terms and then solve for y. 4y = 160 4y = 160 y = 40$^{\circ}$ We can now solve for each individual angle measure. The angle labelled 3y really equals (3 x 40) = 120$^{\circ}$. The angle labelled (y - 15) really equals (40 - 15) = 25$^{\circ}$.
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