Why Use 110 Degrees Instead of 70 for F2 Components?

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Using 110 degrees instead of 70 degrees for calculating the F2 components is crucial due to the orientation of the angles with respect to the x-axis. When using 70 degrees, the cosine value projects onto the negative x-axis, leading to potential confusion with signs. In contrast, using 110 degrees directly accounts for the negative sign in the cosine calculation, simplifying the interpretation of results. The discussion highlights that the diagram may misrepresent the angle relationships, suggesting that the angle should be considered from the positive x-axis to F2. Overall, clarity in angle representation is essential for accurate component analysis.
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Homework Statement
why taking angle 110 instead of 70 for finding the magnitude of F2 in X and Y components?
Relevant Equations
F2x = 150.cos 110 = - 51.30, F2y = 150.sin 110 = 140.95
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F1x = 120. Cos 300 = 103.92 ; F1y = 120.sin 300 = 60N
F2x = 150.cos 110 = - 51.30, F2y = 150.sin 110 = 140.95

why taking angle 110 instead of 70 for finding the magnitude of F2 in X and Y components?
 
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If you take angle ##70°## insteady of ##110°##, do you get negative value for ##F_{2x}## component (as it should be according to the diagram)?
 
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If you are interested in just the magnitude of ##F_{2}## it doesn't matter which angle you take. That's because
##F_2=\sqrt{F_2^2\sin^2(\theta)+F_2^2\cos^2(\theta)}.## Note that ##\cos(110^o)=-\cos(70^o)## and when you square, the negative sign drops out.
 
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I can't let this go. The diagram is very misleading. ?:)

The diagram shows the angular difference between F_1 and F_2 as 110^o. But that doesn't make any sense. That would make the total angle from the positive x-axis to the negative x-axis as 210^o. But that's impossible, since the x-axis is a straight line, and all angles that span a straight line are 180^o. So first and foremost, that needs to be fixed.

From here on out, I'm assuming that the 110^o angle spans from the positive x-axis to F_2 (not F_1 to F_2 as is it shown in the original post).

From there, just follow @kuruman's and @lomidrevo's advice. If you use the 70^o angle, which is the angle with respect to the negative x-axis, then the \cos 70^o will be projected along the negative x-axis. That's a tidbit you need to keep in the back of your mind when interpreting the result. However, if you use the 110^o angle, which is the angle with respect to the positive x-axis, the \cos 110^oresult is already negative so it takes care of the negative sign automatically.
 
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collinsmark said:
From here on out, I'm assuming that the ##110^o## angle spans from the positive x-axis to ## F_2## (not ##F_1## to ##F_2## as is it shown in the original post).
That was my interpretation of the situation. My reading was that OP was questioning whether the difference between ##\cos(70^o)## and ##\cos(110^o)## is more than just an algebraic sign.
 
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