1. ## Geometry help

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

2. Originally Posted by ur5pointos2slo
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
For the member XT, use the Pythagorean theorem.
The altitude of the truss is 2.5m.
The horizontal leg of the imaginary right triangle whose hypotenuse is Xt is
= (half of VWXY) minus XY
= 20/2 -20/3
= 3.33333 m.

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For the member XZ.

By geometry alone, it is a long solution. Can you use trigonometry?
If yes, then,
let angle TYX be angle Y.

tanY = (2.5) /(20/2) = 0.25
So, angle Y = arctan(0.25) = 14.03624 dgrees.

In right triangle XZY,
sinY = (XZ) / (XY)

3. Thanks. Why would you use 20/2 for the XTY Triangle?

4. Originally Posted by ur5pointos2slo
Thanks. Why would you use 20/2 for the XTY Triangle?
XTY triangle? I never used the XTY triangle.

You mean the tanY = 2.5 / (20/2)?
That is not on the XTY triangle.
IT is on the right triangle whose vertical leg is the altitude of the whole truss, and whose horizontal leg is half of VWXY, and whose hypotenuse is TY.

5. I see now. I actually meant to say ANGLE TYX . Thanks so much.