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

`mu = 0.5, N= 2mg (sin theta - mu cos theta)`
`mg cos_(x) = 2mg (sin_(x) - mu cos_(x))`
`cos_(x) = 2cos_(x) (tan_(x) - mu)`
`1/2 = tan_(x) - 1/2 rArr tantheta =1 rArr theta = 45^(@)`