Geometry/Trig


by ur5pointos2sl
Tags: geometry or trig
ur5pointos2sl
ur5pointos2sl is offline
#1
Sep2-08, 11:27 AM
P: 96
A simple roof truss design is shown in figure 2.10( i have attached) The lower section, VWXY is made from three equal length segments. UW and XZ are perpendicular to VT and TY, respectively. If VWXY is 20m and the height of the truss is 2.5m, determine the lengths of XT and XZ.

Ok It has been years since I have taken geometry and am looking for a little guidance. Any theorems or help to get this problem started in the right direction would be great. I do not want full solutions that would be defeating the purpose :)
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HallsofIvy
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#2
Sep2-08, 01:10 PM
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VWXY is 20m long and divided into 3 equal segments so each segment has length 20/3 m. If you draw a perpendicular from T to VWXY, it will bisect WX at, say S, and so forms a right triangle with legs of length (20/3)/2= 10/3 and 2.5 m. You can use the Pythagorean theorem to find the length of the hypotenuse XT: XT2= (10/3)2+ 2.52.

Similarly, that perpendicular forms right triangle STY having legs of length 2.5m and 10m (half of the 20m length of VWXY). TY2= 2.52+ 10[/sup]. Call that x.

Finding XZ is a little harder. Let u= length of TZ. Then length of ZY is x- u. You now have two right triangle with leg XZ. The first right triangle is XTZ which has hypotenuse
XT and legs TZ and XZ: TX2= XZ2+ u2. The other right triangle is XZY which has hypotenuse 10/3 m and legs XZ and TY- u: (10/3)2= XZ2+ (TY- u)2. Since you have calculated TX and TY above that gives two equations to solve for XZ and u.
Diffy
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#3
Sep2-08, 01:16 PM
P: 443
Edit: Deleting my solutions, per OP request. *sorry*

ur5pointos2sl
ur5pointos2sl is offline
#4
Sep2-08, 07:11 PM
P: 96

Geometry/Trig


Quote Quote by HallsofIvy View Post
Finding XZ is a little harder. Let u= length of TZ. Then length of ZY is x- u. You now have two right triangle with leg XZ. The first right triangle is XTZ which has hypotenuse
XT and legs TZ and XZ: TX2= XZ2+ u2. The other right triangle is XZY which has hypotenuse 10/3 m and legs XZ and TY- u: (10/3)2= XZ2+ (TY- u)2. Since you have calculated TX and TY above that gives two equations to solve for XZ and u.
Ok maybe I am missing something. I can solve to get TY to be approximately 10.31. But after that I cannot see how to solve for TZ and ZY. There is always 2 things left unsolved and we all know you cant solve for 2 different things without at least one of them. I am a little confused. I understand the problem setup and where you are trying to go though just seems to be missing some important numbers to help solve.
HallsofIvy
HallsofIvy is offline
#5
Sep3-08, 03:34 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886
You can, as I showed, use triangles XZT and XZY to set up two equations for those two unknown numbers.


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