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 11.2 - The Cosine Ratio and Applications - Exercises - Page 511: 6

Answer

cos α = $\frac{2}{sqrt of 13}$ , cos β = $\frac{3}{sqrt of 13}$

Work Step by Step

Step 1: By Pythagoras theorem $b^{2}$ +$3^{2}$ = square of ($\sqrt 13$) $b^{2}$ + 9 = 13 $b^{2}$ = 13 -9 = 4 b = $\sqrt 4$ = 2 Step 2: cos α =$\frac{ lengthofadjacent}{lengthofhypotenuse}$ cos α = $\frac{2}{sqrt of 13}$ Similarly cos β = $\frac{ lengthofadjacent}{lengthofhypotenuse}$ cos β = $\frac{3}{sqrt of 13}$ Therefore cos α = $\frac{2}{sqrt of 13}$ , cos β = $\frac{3}{sqrt of 13}$
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