Question

Find the magnitude of resultant vector of two vectors a and b of magnitude 7 and 4 respectively if angle between them is 51.3°. What is the sine of angle between r and a

Answer 1

Given,

|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 × 0.611

∴ |R|² = 65 + 35.01

∴ |R|² = 100.01

∴ |R| = 10.0007 ≈ 10

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

We know tanα = BSinθ/(A + BCosθ)

∴ tanα = 4Sin51.3/(7 + 4Cos51.3)

∴ tanα = 4 × 0.78/(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.

Hope it helps.

Answer 2

Answer:

|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 × 0.611

∴ |R|² = 65 + 35.01

∴ |R|² = 100.01

∴ |R| = 10.0007 ≈ 10

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

We know tanα = BSinθ/(A + BCosθ)

∴ tanα = 4Sin51.3/(7 + 4Cos51.3)

∴ tanα = 4 × 0.78/(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