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It is known that A + R = B and R perpendicular A . The angle between A and B is

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7 months ago

```							Magnitude of resultant of A and B is given by,R = √(A² + B² + 2ABcosФ) According to question ,Magnitude of resultant = half of Magnitude of Be.g., √(A² + B² + 2A.BcosФ) = B/2Taking square both sides,A² + B² + 2A.BcosФ = B²/4A² + 2ABcosФ + 3B²/4 = 0 --------(1) Also, A and R is perpendicular upon each other ,e.g., A.(A + B) = 0A.A + A.B = 0A² + A.BcosФ = 0cosФ = - A/B , put it in equation (1)A² + 2A.B(-A/B) + 3B²/4 = 0A² - 2A² + 3B²/4 = 0A = √3B/2Now, cosФ = -A/B = -√3B/2B = -√3/2cosФ = cos150° ⇒Ф = 150°Hence angle between A and B = 150°
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7 months ago
```							Dear studentAccording to question ,Magnitude of resultant = half of Magnitude of Be.g., √(A² + B² + 2A.BcosФ) = B/2Taking square both sides,A² + B² + 2A.BcosФ = B²/4A² + 2ABcosФ + 3B²/4 = 0 --------(1)Also, A and R is perpendicular upon each other ,e.g., A.(A + B) = 0A.A + A.B = 0A² + A.BcosФ = 0cosФ = - A/B , put it in equation (1)A² + 2A.B(-A/B) + 3B²/4 = 0A² - 2A² + 3B²/4 = 0A = √3B/2Now, cosФ = -A/B = -√3B/2B = -√3/2cosФ = cos150° ⇒Ф = 150°Hence angle between A and B = 150°
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7 months ago
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