F1 = 1150 N, F2 = 875 N and F3 = 1450 N are applied at the same point. The angle between F1 and F2 is 90 degrees and the angle between F2 and F3 is 120 degrees. F2 is between F1 and F3. Find the resultant. the answer is 190 N and 126.3 degrees.

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F1 = 1150 N, F2 = 875 N and F3 = 1450 N are applied at the same point. The angle between F1 and F2 is 90 degrees and the angle between F2 and F3 is 120 degrees. F2 is between F1 and F3. Find the resultant. the answer is 190 N and 126.3 degrees.

Physics
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|dw:1437436008915:dw|
This is my work: Fx components: 1150 N + (1450 cos 210) = -105.7368 Fy compnents: 875 N + (1450 sin 210) = 150 \[R= \sqrt{105.7368^{2}+ 105^{2}} = 183.52 N \]
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@IrishBoy123, the back of the book says the answer is 190 N and 126.3 degrees. I am not getting that answer
sounds like your diagram is wrong so we could try some reverse engineering....
can you help me draw a new one then? or just go off the text?
i drew it differently and got your answer again!! let me see if there is another way you clearly know your stuff so it is annoying that you have to guess the diagram....
okay
|dw:1437492586634:dw| i wrote a little script to find the angle \( \theta\) as per my diagram that gives a net force of 190N The output: HIT!! angle = 225.21 F = 190.0 net angle= -39.53 HIT!! angle = 240.25 F = 189.9 net angle= -34.99 so it looks to me like you get a net 190N at \( \theta \) = 225.2 or 240.25 and the angle of that 190N is -39 or -35....... IOW i think the book is wrong, if that helps at all.... [the script is attached if you are interested]
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