## Elementary Geometry for College Students (6th Edition)

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