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120o
Let R be the initial resultant of forces P and Q.
Then, using the law of cosines,
R² = P² + Q² + 2PQcosθ
R² = (2F)² + (3F)² + 2(2F)(3F)cosθ
R² = 4F² + 9F² + 12F²cosθ
R² = 13F² + 12F²cosθ
If the force Q is doubled (becomes 6F), the new resultant R' is doubled (becomes 2R).
(2R)² = P² + (2Q)² + 2P(2Q)cosθ
4R² = (2F)² + (6F)² + 2(2F)(6F)cosθ
4R² = 4F² + 36F² + 24F²cosθ
4R² = 40F² + 24F²cosθ
Substitute R² = 13F² + 12F²cosθ:
4(13F² + 12F²cosθ) = 40F² + 24F²cosθ
52F² + 48F²cosθ = 40F² + 24F²cosθ
12F² = -24F²cosθ
cosθ = -12F²/24F²
cosθ = -1/2
θ = 120°