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DrummerVim

http://gyazo.com/ecaceb89b3a014abf50e53dbeaebf4ef.png?1354481207 I can't get my head around this question. I've got that SR=PRtan35 and SR=QRtan40, which looks like it could help, but I can't figure it out. Please help? :)

  • one year ago
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  1. jim_thompson5910
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    let h = height of pole sin(40) = h/QS QS*sin(40) = h QS = h/sin(40) sin(35) = h/PS PS*sin(35) = h PS = h/sin(35)

    • one year ago
  2. jim_thompson5910
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    each triangle PRS and QRS are right triangles so we can say PR^2 + RS^2 = PS^2 PR^2 + h^2 = PS^2 PR^2 + h^2 = (h/sin(35))^2 --------------------- QR^2 + RS^2 = QS^2 QR^2 + h^2 = QS^2 QR^2 + h^2 = (h/sin(40))^2

    • one year ago
  3. jim_thompson5910
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    Now use what you found SR=PRtan35 and SR=QRtan40 PRtan35 = QRtan40 PR = QR*tan(40)/tan(35)

    • one year ago
  4. jim_thompson5910
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    Since PR = QR*tan(40)/tan(35) we go from PR^2 + h^2 = (h/sin(35))^2 to ( QR*tan(40)/tan(35) )^2 + h^2 = (h/sin(35))^2 if you let x = QR and y = h, then we get this equation ( x*tan(40)/tan(35) )^2 + y^2 = (y/sin(35))^2

    • one year ago
  5. jim_thompson5910
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    again, let x = QR and y = h to go from QR^2 + h^2 = (h/sin(40))^2 to x^2 + y^2 = (y/sin(40))^2

    • one year ago
  6. jim_thompson5910
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    so you now have these 2 equations (with 2 unknowns) ( x*tan(40)/tan(35) )^2 + y^2 = (y/sin(35))^2 x^2 + y^2 = (y/sin(40))^2

    • one year ago
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