Elementary Geometry for College Students (7th Edition)

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

Chapter 1 - Review Exercises - Page 75: 28


m$\angle$EFH = 67.5$^{\circ}$

Work Step by Step

We know that the measure of $\angle$EFG is 90$^{\circ}$ because we are given that it is right. Using the Angle Addition Postulate, m$\angle$EFH+ m$\angle$HFG=m$\angle$EFG. We are also given that m$\angle$EFH = 3*m$\angle$HFG and m$\angle$HFG=2$x$-6 Using the AAP, it can be concluded that 4*m$\angle$HFG = m$\angle$EFG With substitution, 4*m$\angle$HFG = 90$^{\circ}$ Then, m$\angle$HFG =(90/4)$^{\circ}$, or 22.5$^{\circ}$ Using substitution once more, m$\angle$EFH=3*(22.5$^{\circ}$) Finally, after multiplying by 3, m$\angle$EFH=67.5$^{\circ}$
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