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Explain in detail the triangle law of addition.

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#### Solution

Let us consider two vectors `vecA` and `vecB` as shown in the figure. To find the resultant of the two vectors we apply the triangular.

**Law of addition as follows:**

present the vectors A and by the two adjacent sides of a triangle taken in the same order. Then the result is given by the third side of the triangle as shown in the figure.

**Head and tail of vectors**

**Triangle law of addition**

To explain further, the head of the first vector `vecA` is connected to the tail of the second vect `vecB` Let O he the angle between `vecA` and `vecB`. Then `vecR` is the resultant vector connecting the tail of the first vector `vecA` to the head of the second vector `vecB` The magnitude of `vecR`. (resultant) given geometrically by the length of (OQ) and the direction of the resultant vector is the angle between `vecR`. and `vecA`. Thus we write

`vecR = vecA + vec(B) vec(OQ) = vec(OP) + vec(PQ)`

**1. Magnitude of resultant vector:**

The magnitude and angle of the resultant vector are determined by using the triangle law of vectors as follows. From the figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right-angled triangle.

**Resultant vector and its direction by triangle law of addition**

From the figure, let R is the magnitude of the resultant of `vecA` and `vecB`.

cos θ = `"AN"/"B"` ∴ AN = B cos θ and sinθ = `"BN"/"B"` ∴ BN = B sinθ

For ∆ OBN, we have OB^{2} = ON^{2} + BN^{2}

⇒ R^{2} = (A + B cos θ)^{2} + (B sinθ)^{2}

⇒ R^{2} = A^{2} + B^{2} cos^{2}θ + 2ABcosθ B^{2} sin^{2}θ

⇒ R^{2} = A^{2} + B^{2}(cos^{2}θ + sin^{2}θ) + 2AB cos θ

⇒ R^{2} = `sqrt(A^2 + B^2 + 2ABcostheta)`

**2. Direction of resultant vectors:**

If 0 is the angle between `vecA` and `vecB` then,

`|vecA + vecB| = sqrt(A^2 + B^2 + 2ABcostheta)`

If R makes an angle α with `vecA`, then in AOBN,

tan α = `"BN"/"ON" = "BN"/("OA + AN")`

tan α = `(Bsintheta)/(A + Bcostheta) ⇒ α = tan^-1((Bsintheta)/(A + Bcostheta))`

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