Two forces are acting at an angle of 120 degree the bigger force is 14 newton and the resultant perpendicular to the
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Suppose the forces are P and Q. Let the angle between the forces be α.
|P| = |Q|
Let the resultant be R and the resultant makes an angle β with Q.

Now,
tanβ = (P sinα)/(Q + P cosα)
=> tanβ = (P sinα)/(P + P cosα)
=> tanβ = (sinα)/(1 + cosα) = tan(α/2)
=> β = α/2 …………………..(1)
Again,
tan(β/2) = (P/2)(sinα)/[Q + (P/2)cosα]
=> tan(β/2) = (P/2)(sinα)/[Q + (P/2)cosα]
=> tan(β/2) = (1/2)(sinα)/[1 + (1/2)cosα]
=> tan(β/2) = (sinα)/[2 + cosα]
(1) => tan(α/4) = (sinα)/[2 + cosα]
=> [sin(α/2)]/[1 + cos(α/2)] = (sinα)/[2 + cosα]
=> [sin(α/2)]/[1 + cos(α/2)] = [2sin(α/2)cos(α/2)]/[2 + {2cos2(α/2) - 1}]
=> 1/[1 + cos(α/2)] = [2cos(α/2)]/[2cos2(α/2) + 1]
cos(α/2) = x
So,
1/(1 + x) = 2x/(2x2 + 1)
=> 2x2 + 1 = 2x + 2x2
=> x = ½
So,
cos(α/2) = ½ = cos60
=> α = 120o
This is the angle between the forces.
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