The velocity ratio of an inclined plane where angle of inclination in θ is

Explanation

Consider an inclined plane shown below, as the effort moves along OB, the load moves is lifted up through a vertical height AB

V.R = \(\frac{\text{distance moved by effort}}{\text{distance moved by load}}\) = \(\frac{OB}{AB}\)

But Sinθ = \(\frac{opp}{hyp}\) = \(\frac{AB}{OB}\)

therefore; = \(\frac{OB}{AB}\) = \(\frac{1}{\sin \theta}\)

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