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I would like to plot the 'power triangle' in Mathematica, is there a way to do it?

Exactly the way it is presented in this picture from here:

enter image description here

But I've to can change the values in the triangle


As far as I got know:

Show[Graphics[{FaceForm[
    RGBColor[0.5866666666666667, 2/3, 0.5866666666666667]], 
    Triangle[{{0, 0}, {1, 0}, {1, 1}}]}], 
    PlotLabel -> HoldForm[Vermogensdriehoek], 
    LabelStyle -> {GrayLevel[0]}]

I'm stuck here, my axes has to be a arrow and I've to get the text in the right place, any help?

(Vermogensdriehoek is Dutch for Power Triangle)

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    $\begingroup$ Is there any particular reason to think there isn't a way to do this? I mean, what about it has given you problems, specifically? $\endgroup$ – Oleksandr R. Sep 17 '15 at 16:59
  • $\begingroup$ They (someone I am working with) asked me to make it in Mathematica and I've no idea how to deal with it, that's why I decide to ask it here $\endgroup$ – Jan Sep 17 '15 at 17:03
  • $\begingroup$ Please understand that this isn't a code-it-for-me service! $\endgroup$ – Dr. belisarius Sep 17 '15 at 17:05
  • $\begingroup$ Hints are also welcome, bro ;) I hope I am not offending you, if I did I'm sorry $\endgroup$ – Jan Sep 17 '15 at 17:06
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 Manipulate[
 Graphics[
 {FaceForm[RGBColor[0.587, 2/3, 0.587]], 
 Triangle[{{0, 0}, r {Cos[ϕ], 0}, r {Cos[ϕ], Sin[ϕ]}}],
 Circle[{0, 0}, .1 r, {0, ϕ}],
 Text[Style["ϕ", 14, Italic], 
    .15 r {Cos[ϕ/2], Sin[ϕ/2]}],
 Text[Style["Power\n triangle"], 
    r {Cos[ϕ]/2, Sin[ϕ]/4}]},
  PlotRange -> {{0, 2}, {0, 2}},
  Axes -> True,
  ImageSize-> 600,
  AxesLabel -> {"P (Real Power)", "I (Apparent Power)"}],
 {{ϕ, .2}, 0, π/2},
 {{r, 1}, .5, 2}
 ]
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  • $\begingroup$ Can you use my code and make it bigger, with your information? $\endgroup$ – Jan Sep 17 '15 at 18:43
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    $\begingroup$ You certainly have the start you requested. I think you're on your own for the refinements. $\endgroup$ – David G. Stork Sep 17 '15 at 18:52
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    $\begingroup$ @JanEerland You see? That was what I was referring to in my comments under your question. Mathematica is too vast and complex to try to find your way out of a problem with a simple hint and without spending a lot of your time afterwards $\endgroup$ – Dr. belisarius Sep 17 '15 at 20:15
  • $\begingroup$ Sure I see! The only thing I wanna know now is how to make that half Phi circle in my code: Show[Graphics[{FaceForm[RGBColor[0.586667, 2/3, 0.586667]], Triangle[{{0, 0}, {1, 0}, {1, 1}}]}, {Axes -> True, PlotLabel -> HoldForm[Vermogensdriehoek], LabelStyle -> {GrayLevel[0]}}], AxesLabel -> {HoldForm[Re -> P = UICos[[CurlyPhi]]], HoldForm[Im -> Q = UISin[[CurlyPhi]]]}, PlotLabel -> None] $\endgroup$ – Jan Sep 17 '15 at 20:17
  • $\begingroup$ But what you call the "half Phi circle" (actually, the full Phi arc) is implemented in my code, as you can surely see when you run it. $\endgroup$ – David G. Stork Sep 17 '15 at 21:14

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