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