The tension of the right-hand string divided by the tension of the left-hand string is given by A) sin $\theta$₂ / sin $\theta$₁ B) cos $\theta$₁ / cos $\theta$₂ C) cos $\theta$₂/ cos $\theta$₁ D) sin $\theta$₁ / sin $\theta$₂ E) (mg - T₁ sin $\theta$₁) / (T₁ $\times$ sin $\theta$₂)

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The Answer of The tension of the right-hand string divided by the tension of the left-hand string is given by A) sin $\theta$₂ / sin $\theta$₁ B) cos $\theta$₁ / cos $\theta$₂ C) cos $\theta$₂/ cos $\theta$₁ D) sin $\theta$₁ / sin $\theta$₂ E) (mg - T₁ sin $\theta$₁) / (T₁ $\times$ sin $\theta$₂)

The Tension of the Right-Hand String Divided by the Tension θ\thetaθ

The Tension of the Right-Hand String Divided by the Tension $\theta$