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