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
Answers
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:
|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