Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 11 - Section 11.2 - The Cosine Ratio and Applications - Exercises - Page 503: 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 = $\frac{3}{sqrt of 13}$
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.