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 11 - Section 11.4 - Applications with Acute Triangles - Exercises - Page 527: 9b

Answer

γ = 90°

Work Step by Step

By using theorem 11.4.3 $c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab cos γ $5^{2}$ = $3^{2}$ + $4^{2}$ - 2*3*4* cos γ 25 = 9 + 16 - 24cos γ 25 = 25 - 24cos γ 25 - 25 = - 24cos γ 0 = - 24cos γ cos γ = 0 γ = $cos^{-1}$(0) γ = 90°
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.