Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 5 - Section 5.6 - Segments Divided Proportionally - Exercises - Page 268: 18

Answer

$\angle$PNQ=45$^{\circ}$

Work Step by Step

$\frac{NP}{MN}$=$\frac{PQ}{MQ}$ $\frac{4}{8}$=$\frac{3}{6}$ $\frac{1}{2}$=$\frac{1}{2}$ QN bisects angle N because the corresponding segments of the triangle are proportional $\angle$N=180-$\angle$M-$\angle$P $\angle$N=180-63-27 $\angle$N=90 2$\angle$PNQ=$\angle$N 2$\angle$PNQ=90 $\angle$PNQ=45
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