Two forces 3 N and 2 N are at angle  such that theresultant is R. The first force is now increased to 6N and the resultant become 2R. The value of  is

The problem involves vector addition and trigonometry. Here are the steps to solve it:

Step 1: The resultant of two forces can be found using the formula: R = sqrt((F1^2) + (F2^2) + 2F1F2*cos(θ)), where F1 and F2 are the magnitudes of the forces and θ is the angle between them.

Step 2: Given that the forces are 3N and 2N and the resultant is R, we can write the equation as: R = sqrt((3^2) + (2^2) + 232*cos(θ)).

Step 3: Simplifying the equation gives: R = sqrt(13 + 12cos(θ)).

Step 4: It's also given that when the first force is increased to 6N, the resultant becomes 2R. We can write this equation as: 2R = sqrt((6^2) + (2^2) + 262*cos(θ)).

Step 5: Simplifying this equation gives: 2R = sqrt(40 + 24cos(θ)).

Step 6: Now, we have two equations with the same variable θ. We can solve for θ by equating the two equations: sqrt(13 + 12cos(θ)) = sqrt(40 + 24cos(θ)).

Step 7: Squaring both sides to get rid of the square root gives: 13 + 12cos(θ) = 40 + 24cos(θ).

Step 8: Solving for cos(θ) gives: cos(θ) = -27/12 = -2.25, which is not possible as the value of cos(θ) lies between -1 and 1.

There seems to be a mistake in the problem as it leads to an impossible value for cos(θ). Please check the problem and try again.