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in Physical World and Measurement by (15 points)
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Find the magnitude of resultant vector R of two vectors A and B of magnitudes 7 and 4 unit if angle between them is 51.3 degree

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1 Answer

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We have

|A| = 7 and |B| = 4 and θ = 51.3°.

Using the formula,

|R|² = |A|² + |B|² + 2|A||B|Cosθ

|R|² = 49 + 16 + 2 × 7 × 4 × Cos51.3

|R|² = 65 + 56 × 0.625

∴ |R|² = 65 + 35

∴ |R|² = 100

∴ |R| = 10 ≈ 10

Let the angle between the Vector A and Vector R be α. Then we need to Find Sinα.

Cos 51.3 = 5/8 = 0.625

Sin 51.3 = √39/8 = 0.781

We know tanα = BSinθ/(A + BCosθ)

∴ tanα = 4Sin51.3/(7 + 4Cos51.3)

∴ tanα = 4 × 0.781/(7 + 4 × 0.625)

∴ tanα = 3.12/9.5

∴ tanα = 0.328

∴ α = tan⁻¹(0.328)

∴ α = Sin⁻¹[(0.328)/√(1 + 0.328²)]

∴ α = Sin⁻¹[0.328/1.107]

∴ α = Sin⁻¹[0.296]

∴ Sinα = 0.296

Hence, the Sine of angle between A and R is 0.296.

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