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2 years ago

Eshan
2095 Points
							Dear student,Since the vectors form a right angled triangle,$\dpi{80} Bsin\theta=R=\dfrac{B}{\sqrt{2}}$$\dpi{80} \implies sin\theta=\dfrac{1}{\sqrt{2}}$$\dpi{80} \implies \theta=45^{\circ}$

2 years ago
prateek
14 Points
							We have the vertical component of vector B = B sinθThe vertica component will be equal to the resultant vector. B Sinθ = B/√2we get   Sinθ= 1/√2 Angle b/w vectors is

2 years ago
prateek
14 Points
							  We have the vertical component of vector B = B sinθThe vertica component will be equal to the resultant vector. B Sinθ = B/√2we get   Sinθ= 1/√2 Angle b/w vectors is 45°

2 years ago
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### Course Features

• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions