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Question no. 22 solution please . How i get the answer

Question no. 22 solution please . How i get the answer

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Grade:12th pass

3 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Dear student,

Since the vectors form a right angled triangle,Bsin\theta=R=\dfrac{B}{\sqrt{2}}

\implies sin\theta=\dfrac{1}{\sqrt{2}}
\implies \theta=45^{\circ}
prateek
14 Points
5 years ago
We have the vertical component of vector B = B sinθ
The vertica component will be equal to the resultant vector.
 B Sinθ = B/√2
we get   
Sinθ= 1/√2 
Angle b/w vectors is 
 
prateek
14 Points
5 years ago
 
 
We have the vertical component of vector B = B sinθ
The vertica component will be equal to the resultant vector.
 B Sinθ = B/√2
we get   
Sinθ= 1/√2 
Angle b/w vectors is 45°

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