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 - Section 11.4 - Applications with Acute Triangles - Exercises - Page 529: 40

Answer

$c^{2}= a^{2} +b^{2}- 2\times a \times b \times 0= c^{2}= a^{2} +b^{2}$

Work Step by Step

If $\gamma = 90^{\circ}$, then $\cos \gamma =0$ Therefore, if we replace $\cos \gamma$ by 0 in the equation, we get $c^{2}= a^{2} +b^{2}- 2\times a \times b \times 0$ $ c^{2}= a^{2} +b^{2}$ And that is the Pythagorean Theorem
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