Find the values of $\theta$ and $p$, if the equation $x \cos \theta+y \sin \theta=p$ is the normal form of the line $\sqrt{3} x+y+2=0$.
The equation of the given line is $\sqrt{3} x+y+2=0$.
This equation can be reduced as
$\sqrt{3} x+y+2=0$
$\Rightarrow-\sqrt{3} x-y=2$
On dividing both sides by $\sqrt{(-\sqrt{3})^{2}+(-1)^{2}}=2$, we obtain
$-\frac{\sqrt{3}}{2} x-\frac{1}{2} y=\frac{2}{2}$
$\Rightarrow\left(-\frac{\sqrt{3}}{2}\right) x+\left(-\frac{1}{2}\right) y=1$ $\ldots(1)$
On comparing equation (1) to $x \cos \theta+y \sin \theta=p$, we obtain
$\cos \theta=-\frac{\sqrt{3}}{2}, \sin \theta=-\frac{1}{2}$, and $p=1$
Since the values of $\sin \theta$ and $\cos \theta$ are negative, $\theta=\pi+\frac{\pi}{6}=\frac{7 \pi}{6}$
Thus, the respective values of $\theta$ and $p$ are $\frac{7 \pi}{6}$ and 1
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