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 11.1 - The Sine Ratio and Applications - Exercises - Page 504: 6

Answer

sin α=$\frac{3}{\sqrt {13}}$ sin β= $\frac{2}{\sqrt {13}}$

Work Step by Step

In given right triangle, applying Pythagorean theorem- $(\sqrt {13})^{2} = b^{2} + 3^{2}$ Therefore- $b^{2} = (\sqrt {13})^{2} - 3^{2}$ = 13 - 9 = 4 a = $\sqrt {4}$ = 2 sin α=$\frac{opposite}{hypotenuse}=\frac{3}{\sqrt {13}}$ sin β=$\frac{opposite}{hypotenuse}=\frac{b}{\sqrt {13}}$ = $\frac{2}{\sqrt {13}}$
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