I need someone to explain this vector question's answer

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Vectors A and B, both with magnitudes of 57, sum to the vector 16.3j, leading to the calculation of the angle between them. The correct approach involves using the cosine function, resulting in an angle of 81.8º, which means the angle between the vectors is 163.6º. The discussion emphasizes the importance of visualizing the problem with a diagram, as the x components of the vectors cancel each other out. Misapplication of the sine function was clarified, highlighting that it applies to the complementary angle in the context of the triangle formed by the vectors. Ultimately, drawing a diagram is crucial for understanding vector relationships in two dimensions.
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Homework Statement



Vectors A and B have equal magnitudes of 57. If the sum of A and B is the vector 16.3j , determine the angle?



The Attempt at a Solution


I have already gotten the answer
2•57cosθ = 16.3
cosθ = 0.143
θ = 81.8º
twice that is the angle between them, or 163.6º
but at first i did 2*57sinθ=16.3 since we have the y component and as far as I know y=rsinθ
why is the right answer cos?
 
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madinsane said:
Vectors A and B have equal magnitudes of 57. If the sum of A and B is the vector 16.3j , determine the angle?
So draw the diagram, in 2 dimensions. Since the vectors are of equal length, and their x components cancel, the second vector must be a reflection of the first in the y axis.

as far as I know y=rsinθ
Working without a sketch of what you are dealing with is a dangerous thing! If you refer to your diagram, you will see that y=rsinθ only if θ is the other acute angle in the triangle. So rsinθ will find you the complementary angle of what you are needing here.

It all becomes clear when you draw a diagram.
 
NascentOxygen said:
So draw the diagram, in 2 dimensions. Since the vectors are of equal length, and their x components cancel, the second vector must be a reflection of the first in the y axis.


Working without a sketch of what you are dealing with is a dangerous thing! If you refer to your diagram, you will see that y=rsinθ only if θ is the other acute angle in the triangle. So rsinθ will find you the complementary angle of what you are needing here.

It all becomes clear when you draw a diagram.
Thank you! you were right! I got it now
 
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