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Equation solving

  1. Apr 17, 2007 #1
    Can anyone tell me how to solve....

    400sinA + 500cosA = 600

    Just doin some statics problems and for one of 'em I needed to find A. Can anyone tell me how its done.

  2. jcsd
  3. Apr 17, 2007 #2
  4. Apr 18, 2007 #3


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    First you need to know that sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B) for any A, B. Because we obviously can't have cos(B)= 400 and sin(B)= 500 (sin and cos values are always <= 1), multiply that entire equation by C: Csin(A+ B)= Csin(A)cos(B)+ Ccos(A)sin(B). You want to find C and B so that C cos(B)= 400 and C sin(B)= 500. Since sin2(B)+ cos2(B)= 1, C2 cos2(B)+ C2sin2(B)= C2= (400)2+ (500)2= 160000+ 250000= 410000. C2= 410000 so [itex]C= \sqrt{410000}= 100\sqrt{41}[/itex]. Now you know that [itex]C cos(B)= 100\sqrt{41}cos(B)= 400 so [itex]cos(B)= 4/\sqrt{41}[/itex]. [itex]B= arccos(4/\sqrt{41})[/itex] so we now have
    [tex]100\sqrt{41} sin(A+ arccos(4/\sqrt{41})= 600[/tex]
    [tex]sin(A+ arccos(4/\sqt{41})= 6/\sqrt{41}[/tex]
    so [itex]A+ arccos(4/\sqrt{41})= arcsin(6/\sqrt{41})[/itex] and, finally,
    [tex]A= arcsin(6/\sqrt{41})- arccos(4/\sqrt{41})[/tex]
    According to my calculator, that is about 0.318.
  5. Apr 18, 2007 #4
    Using [tex]sin(x)=\sqrt{1-cox^2(x)}[/tex] I arrived also at a second answer, x=1.0315.
  6. Apr 18, 2007 #5
    I'm not sure your static problem would lead to that equation. Sum of forces in x direction equal 0 and likewise in y. Your equation implies adding an x component and a y component. Once again I don't know what the problem is so I could be wrong.
  7. Apr 18, 2007 #6
    Huh, I think he is adding moments of forces.
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