A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of the resultant downward force along the inclination the angle between the inclined plane and the horizontal is

Explanation

To solve the problem step by step, we will analyze the forces acting on the body sliding down the inclined plane and apply the conditions given in the question.

### Step 1: Identify the forces acting on the body
- The weight of the body \( W = mg \) acts vertically downwards.
- The component of the weight acting parallel to the inclined plane is \( W_{\parallel} = mg \sin \theta \).
- The component of the weight acting perpendicular to the inclined plane is \( W_{\perpendicular} = mg \cos \theta \).
- The normal force \( N \) acts perpendicular to the inclined plane.
- The frictional force \( F_f \) opposes the motion and is given by \( F_f = \mu N = 0.5 N \).
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