physics problem....need some help ??

Question

Imagine a lower leg being exercised. It has a 49-N weight attached to the foot and is extended at an angle Θ with respect to the vertical. Consider a rotational axis at the knee.(a) When Θ = 90.0 degrees, find the magnitude of the torque that the weight creates. (b) At what angle Θ does the magnitude of the torque equal 15N*m?

additional data:
distance from knee to foot: 0.55m

Answer 1

The weight of the weight (as it were) will of course always be in a downwards direction, so we have
τ = 49 (0.55) sin Θ
a) When Θ = 90.0°, sin Θ = 1 so τ = 49(0.55) = 27 Nm.
b) 49 (0.55) sin Θ = 15
=> sin Θ = 15/(49(0.55))
=> Θ = 34°.