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Thread: Trigonometry - finding a length

  1. #1

    Trigonometry - finding a length

    There are 6 points (U,V,W,X,Y,Z) in a 2D space.

    * Point Y lies on line WX
    * Point Z lies on line UX
    * VW and YZ are perpendicular to WX
    * UY and VY are equal length

    VW = 400, XY = 200, YZ = 100 and angle UYV is 20 degrees. Find the length of WY?

    I haven't had much luck tackling this question. Tried right-angle trig, Pythagoras and angle theorems but I cannot get angle UYV into a formula with any of the provided lengths. Could someone point me to the right direction?

    Or at least, can someone prove whether UX and WX are equal or not? (that's the only part I'm missing)
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  2. #2
    Senior Member Peritus's Avatar
    Nov 2007
    1. At first let us find the value of angle UXW:

    using the sin theorem we obtain the following equation:

    YZ / sin(UXW) = XY / sin(90 - UXW) = XY / cos(UXW)

    -> UXW = arctan(0.5) ~ 26.565

    2. now let us apply the sin theorem to triangle XYU (I'll define UY & VY as L)

    L / sin(UXW) = XY / sin(XUY)

    3. we apply the sin theorem again only this time in triangle VYW:
    L / sin(90) = VW / sin (VYW)

    4. the angle VYW can be easily obtained if we observe that angle YZU equals 116.565, ang(ZYU) = 63.435-sin(XUY) -> ang(VYW) = sin(XUY + 6.565)

    thus we obtain two equations in two unknows namely (ang(XUY) and L):

    I. L/sin(26.565) = 200 / sin(XUY)
    II. L = 400 / sin(XUY + 6.565) = 400 / [sin(XUY)cos(6.565) + sin(6.565)cos(XUY)]

    (WY can be easily obtained using the pythagorean theorem after you'll find L)
    I think that u'll be able to do the rest...
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