(a). \(\frac{T}{\sin 90^o}\) = \(\frac{120}{sin 135^o}\) and found T = 169.71N (b) \(\frac{R}{\sin 135^o}\) = \(\frac{120}{\sin 135^o}\) R = 120N

Answer Details

The given problem involves two parts: finding the tension (T) in a rope and the force (R) acting on an object at a certain angle.

In part (a), we are given the weight of an object as 120 N and the angle at which the rope is being pulled as 135 degrees. We use the formula: force = weight / sin(angle) to calculate the tension in the rope. We substitute the given values and calculate T as 169.71 N.

In part (b), we are given the weight of the same object as 120 N and the angle at which the force is acting as 135 degrees. We use the same formula: force = weight / sin(angle) to calculate the force (R) acting on the object. We substitute the given values and calculate R as 120 N.

Therefore, we can conclude that in part (a), we found the tension in the rope when the object was being pulled at an angle of 135 degrees, and in part (b), we found the force acting on the object when the force was being applied at an angle of 135 degrees.