Why Use 110 Degrees Instead of 70 for F2 Components?

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Benjamin_harsh
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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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I can't let this go. The diagram is very misleading. ?:)

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

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

From there, just follow @kuruman's and @lomidrevo's advice. If you use the [itex]70^o[/itex] angle, which is the angle with respect to the negative x-axis, then the [itex]\cos 70^o[/itex] 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 [itex]110^o[/itex] angle, which is the angle with respect to the positive x-axis, the [itex]\cos 110^o[/itex]result 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.