The two forces are acting such that the sum of their magnitude is 18N, the resultant is 228 when they are at 120°. Then the magnitude of the forces are

If F1+F2=18→1 F=228 when θ=120 0⇒F2=F12+F22+2F1F2 cosθF12+F22−F1F2=228→2eq−12 gives F12+F22+2F1F2=324 →3On solvingF1=16N F2=2N

The two forces are acting such that the sum of their magnitude is 18N, the resultant is 228 when they are at 120°. Then the magnitude of the forces are

If F1+F2=18→1 F=228 when θ=120 0⇒F2=F12+F22+2F1F2 cosθF12+F22−F1F2=228→2eq−12 gives F12+F22+2F1F2=324 →3On solvingF1=16N F2=2N

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The two forces are acting such that the sum of their magnitude is 18N, the resultant is $\sqrt{228}$ when they are at $120 ° .$ Then the magnitude of the forces are

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12N, 6N

10N, 9N

13N, 5N

16N, 2N

answer is D.

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Detailed Solution

If $F_{1} + F_{2} = 18 \to 1$

$F = \sqrt{228} when θ = 120^{^{0}}$

$\Rightarrow F^{2} = {F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} cosθ$

${F_{1}}^{2} + {F_{2}}^{2} - F_{1} F_{2} = 228 \to 2$

${(eq - 1)}^{2} gives {F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} = 324 \to 3$

On solving

$F_{1} = 16 N F_{2} = 2 N$

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The two forces are acting such that the sum of their magnitude is 18N, the resultant is 228 when they are at 120°. Then the magnitude of the forces are

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The two forces are acting such that the sum of their magnitude is 18N, the resultant is $\sqrt{228}$ when they are at $120 ° .$ Then the magnitude of the forces are