1. ## Triangle Building Equation

Hi,

I'm looking to create a formula that will solve for side C of a triangle. There's a catch though, so it's not your everyday formula.

Attached is a picture of what the formula needs to solve and what variables will exist.

What makes this tricky is the 65' and 25' angle on one side of the 2'x4'. I've tried making the cuts based off the normal triangle formula but the boards end up short and not meeting flush at he peak.

The blue dotted line is what I'm trying to solve and/or, what the true length of the board should be BEFORE the cuts are made.

2. Originally Posted by kyle4rp
Hi,

I'm looking to create a formula that will solve for side C of a triangle. There's a catch though, so it's not your everyday formula.

Attached is a picture of what the formula needs to solve and what variables will exist.

What makes this tricky is the 65' and 25' angle on one side of the 2'x4'. I've tried making the cuts based off the normal triangle formula but the boards end up short and not meeting flush at he peak.

The blue dotted line is what I'm trying to solve and/or, what the true length of the board should be BEFORE the cuts are made.

First, we will need some other dimension(s) somewhere. Possibly from the rafter midpoint to the outside top (call that A) or from the rafter midpoint to the outside edge (call that B) or ...

Next, look at the top. That 65$^{\circ}$ is not the angle at the top [as I understand the image]. Rather half that (32.5$^{\circ}$) is the angle on the board cut which would give a distance to the inside top (call that A') of A - $\dfrac{7}{2\, cos(32.5^{\circ})}$. Remember that a '2X4' is really only 1.5X3.5. Do the same sort of thing for side associated with B. That will give you better dimensions to work with.

If I have misread the image, please specifically state what the angles are associated with. Oh, and you will still need the A or B or ...

3. Dunno...seems to me you're asking how to cut
the 2by4's so that they "fit" at the top, and so that
they "fit" where they meet the perpendicular 2by4's.