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Q.
If the normal form of the equation $\sqrt{3} x+y-8=0$ is $x \cos \omega+y \sin \omega=p$, then $p$ and $\omega$ respectively are
Straight Lines
Solution:
Given equation is
$\sqrt{3} x+y-8=0 ....$(i)
Dividing Eq. (i) by $\sqrt{(\sqrt{3})^2+(1)^2}=2$, we get
$\frac{\sqrt{3}}{2} x+\frac{1}{2} y=4 \text { or } \cos 30^{\circ} x+\sin 30^{\circ} y=4....$(ii)
Comparing Eq. (ii) with $x \cos \omega+y \sin \omega=p$, we get $p=4$ and $\omega=30^{\circ}$.