Horizontal
Since the inclined plane is inclined at \(30^{\circ}\) to the vertical, then the required angle to the horizontal is \(90^{\circ}-30^{\circ}=60^{\circ}\). See diagram above.
Therefore
\(V \cdot R=\frac{1}{\sin \theta} ; V R=\frac{1}{\sin 60}=\frac{1}{\frac{\sqrt{3}}{2}}\) \(=1 \times \frac{2}{\sqrt{3}}: V R=\frac{2}{\sqrt{3}}\)
by rationalization
\(V R=\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} ; V R=\frac{2 \sqrt{3}}{3}\)EWARE of option A which is 2. This is what you get when Jou use \(\theta\) as \(30^{\circ}\)