## Algebra and Trigonometry 10th Edition

$a \sin \beta \theta+b \cos \beta \theta$
$\tan C=\dfrac{opposite}{adjacent}=\dfrac{b}{a}$ $hypotenuse =\sqrt {a^2+b^2}$ $\cos C=\dfrac{adjacent}{hypotenuse}=\dfrac{a}{\sqrt {a^2+b^2}}$ $\sin C=\dfrac{opposite}{hypotenuse}=\dfrac{b}{\sqrt {a^2+b^2}}$ Now, $\sqrt {a^2+b^2}(\sin \beta \theta \cos C +\cos \beta \theta \sin C)$ or, $\sqrt {a^2+b^2}[\sin \beta \theta (\dfrac{a}{\sqrt {a^2+b^2}}) +\cos \beta \theta (\dfrac{b}{\sqrt {a^2+b^2}})]$ or, $a \sin \beta \theta+b \cos \beta \theta$