Find the equation of the line whose perpendicular distance from the origin is $4$ units and the angle which the normal makes with the positive direction of the $x$-axis is $15^{\circ}$.

(N/A) The normal form of the equation of a line is given by $x \cos \omega + y \sin \omega = p$,where $p$ is the perpendicular distance from the origin and $\omega$ is the angle the normal makes with the positive $x$-axis.
Given $p = 4$ and $\omega = 15^{\circ}$.
We know that $\cos 15^{\circ} = \cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Similarly,$\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.
Substituting these values into the normal form equation:
$x \left( \frac{\sqrt{3} + 1}{2\sqrt{2}} \right) + y \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) = 4$.
Multiplying both sides by $2\sqrt{2}$,we get:
$(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 8\sqrt{2}$.