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resultant force

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Knowing that $P_1=\\var{P1}$kN,  $P_2=\\var{P2}$kN  and  $P_3=\\var{P3}$kN, 

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(a) You can define angles $\\alpha_1$, $\\alpha_2$ and $\\alpha_3$, as shown above.

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Then you can either calculate the angles:

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$\\alpha_1=\\arctan \\dfrac{900}{500}=58.11^\\circ$

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$\\alpha_2=\\arctan \\dfrac{900}{480}=62.93^\\circ$

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$\\alpha_3=\\arctan \\dfrac{600}{800}=36.87^\\circ$

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or calculate the following cosines and sines using their definitions and Pythagoras' Theorem:

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For $\\alpha_1$: 

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$\\cos \\alpha_1=\\dfrac{560}{\\sqrt{560^2+900^2}}=\\dfrac{560}{1060}=0.5283$

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$\\sin \\alpha_1=\\dfrac{900}{\\sqrt{560^2+900^2}}=\\dfrac{900}{1060}=0.8491$

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For $\\alpha_2$: 

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$\\cos \\alpha_2=\\dfrac{480}{\\sqrt{480^2+900^2}}=\\dfrac{480}{1020}=0.4706$

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$\\sin \\alpha_2=\\dfrac{900}{\\sqrt{560^2+900^2}}=\\dfrac{900}{1060}=0.8824$

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For $\\alpha_2$: 

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$\\cos \\alpha_3=\\dfrac{800}{\\sqrt{800^2+600^2}}=\\dfrac{800}{1000}=0.8$

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$\\sin \\alpha_3=\\dfrac{600}{\\sqrt{800^2+600^2}}=\\dfrac{600}{1000}=0.6$

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Since you know these parameters, calculate the components, for example

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$P_{2x}=P_2\\cos \\alpha_2=(\\var{P2}\\;\\text{kN}) 0.4706=\\var{comp[1][0]}\\;\\text{N}$

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$P_{2y}=- P_2\\sin \\alpha_2=(\\var{P2}\\;\\text{kN})0.8824=\\var{comp[1][1]}\\;\\text{N}$

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Now proceed the same way as in the previous question.

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Determine the $x$ and $y$ components of each of the forces shown. 

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Force x-component, kN y-Component, kN
$P_1=${P1} kN[[0]][[1]]
$P_2=${P2} kN[[2]][[3]]
$P_3=${P3} kN[[4]][[5]]
\n

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Determine the $x$ and $y$ components of the resultant force. 

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$R_x=$[[0]], kN

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$R_y=$[[1]], kN

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Determine the magnitude of the resultant force, kN.

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$R=$[[0]]

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Determine the angle (in degrees) that the resultant force forms with axis $x$. Submit a positive angle if the resultant force points up  or negative if down  from the $x$ axis. 

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$\\alpha=$[[0]]

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