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: 5

Answer

$\sin\alpha=\frac{\sqrt3}{\sqrt5}$ $\sin\beta=\frac{\sqrt2}{\sqrt5}$

Work Step by Step

Using Pythagorean Theorem to find $c$. $c=\sqrt {\sqrt2^{2}+\sqrt3^{2}}$ $c=\sqrt {2 + 3}$ $c=\sqrt {5}$ $\sin\alpha=\frac{opposite}{hypotenuse}=\frac{\sqrt3}{c}=\frac{\sqrt3}{\sqrt5}$ $\sin\beta=\frac{opposite}{hypotenuse}=\frac{\sqrt2}{c}=\frac{\sqrt2}{\sqrt5}$
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