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Example · Example 2

Q.Two forces of magnitude 5 N and 5 N act at a point, with an angle of 60∘60^\circ between them. Using the triangle law of vector addition, find the magnitude of the resultant force and the angle it makes with each of the two forces.

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Using R=A2+B2+2ABcos⁡θR = \sqrt{A^2+B^2+2AB\cos\theta} with A=B=5A=B=5 N and θ=60∘\theta = 60^\circ: R=25+25+2(5)(5)cos⁡60∘=50+50(0.5)=75=53≈8.66R = \sqrt{25+25+2(5)(5)\cos60^\circ} = \sqrt{50+50(0.5)} = \sqrt{75} = 5\sqrt3 \approx 8.66 N. Since the two forces are equal in magnitude, the resultant bisects the angle between them, so it makes an angle of 30∘30^\circ with each force — confirmed by tan⁡α=Bsin⁡θA+Bcos⁡θ=5sin⁡60∘5+5cos⁡60∘=4.337.5=0.577\tan\alpha = \dfrac{B\sin\theta}{A+B\cos\theta} = \dfrac{5\sin60^\circ}{5+5\cos60^\circ} = \dfrac{4.33}{7.5} = 0.577, giving α=30∘\alpha = 30^\circ. [!ANSWER] Resultant =53≈8.66= 5\sqrt3 \approx 8.66 N, making 30∘30^\circ with each of the two 5 N forces.

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