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Q.Find the magnitude of the resultant of two vectors A⃗\vec{A} and B⃗\vec{B} and angle θ\theta between them.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2026Subjective· 3mImportance★★★★★
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Parallelogram law gives R=A2+B2+2ABcos⁡θR = \sqrt{A^{2}+B^{2}+2AB\cos\theta} and tan⁡ϕ=Bsin⁡θA+Bcos⁡θ\tan\phi = \dfrac{B\sin\theta}{A+B\cos\theta}.

Let A⃗\vec A and B⃗\vec B act at a point with angle θ\theta between them (parallelogram law). Represent A⃗\vec A by OP and B⃗\vec B by OQ; complete the parallelogram OPSQ, so the resultant is OS.

Drop a perpendicular from S to the extension of OP, meeting it at N. Then PN=Bcos⁡θPN = B\cos\theta and SN=Bsin⁡θSN = B\sin\theta.

In right triangle OSN:

OS2=ON2+SN2=(OP+PN)2+SN2.OS^{2} = ON^{2} + SN^{2} = (OP + PN)^{2} + SN^{2}.

R2=(A+Bcos⁡θ)2+(Bsin⁡θ)2.R^{2} = (A + B\cos\theta)^{2} + (B\sin\theta)^{2}.

Expanding:

R2=A2+2ABcos⁡θ+B2cos⁡2θ+B2sin⁡2θ=A2+B2+2ABcos⁡θ.R^{2} = A^{2} + 2AB\cos\theta + B^{2}\cos^{2}\theta + B^{2}\sin^{2}\theta = A^{2} + B^{2} + 2AB\cos\theta. …

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