# How to calculate miter angles for an angled stretcher

See the image below:

When I lay this out in Fusion, I can get the angles but I can't figure out how to calculate it mathematically.

The goal is to have the angled stretcher fully contact the upper 45"-long support and the floor(even though there will be excess to remove.)

I'm looking for a formula for calculating the stretcher angle(35.8 in this case) and the length that the 2x4 would have to be. Here's my method:

Consider a rectangle that is $$a \times b$$. This is the interior of your structure where the thickness of the supporting members is neglected. Then use trig to write the sides of the outlined triangle in terms of $$a$$ and $$\theta$$. In the small triangle at bottom-right, decompose it into two similar right triangles and write the relevant sides lengths in terms of $$a$$, $$b$$, $$w$$ and $$\theta$$.  Now you can represent the length of your beam as the sum of three lengths: \begin{align} \text{Length} &= w\cdot \cot(\theta) + a\cdot \sec(\theta) + \frac{b-a\cdot \tan(\theta)}{\sin(\theta)} \\ &= w\cdot \cot(\theta) + a\cdot \sec(\theta) + b\cdot \csc(\theta) - a\cdot \sec(\theta)\\ &= w\cdot \cot(\theta) + b\cdot \csc(\theta)\end{align}

Now we have the total length of the beam in terms of just $$b$$, $$w$$ and $$\theta$$, so we need another equation in order to find $$\theta$$.

Notice that the pink and yellow triangles are similar, so write an equation involving the ratios of their corresponding sides: $$\frac{b}{a\cdot\tan(\theta)}=\frac{a+w\cdot\csc(\theta)}{a}$$

I asked Wolfram|Alpha to solve this for $$\theta$$ and it gave:

$$\theta = 2 \tan^{-1}\left(\frac{\sqrt{a^2+b^2-w^2}-a}{b+w}\right)$$

Now we can find the angle from $$a$$, $$b$$ and $$w$$, and we can use the angle to find the length of the beam.

I checked this with your numbers: $$a=38$$, $$b=31.75$$ and $$w=3.5$$

This gave $$\theta \approx 35.8265^{\circ}$$ (which slightly concerns me, since your program rounded this down to 35.82) and $$\text{Length} \approx 59.09 \text{in}$$.

Last check: I found a long scrap of plywood in my garage that is 7.75 inches wide. I took it into my kitchen and laid it so it spanned a $$2\times 3$$ rectangle of floor tile. The formula gave 78.43 inches and it measured just over 78.5. Given the spacing of the grout between the tiles, I'm satisfied with this result.

• This is correct. If you want to edit, the actual length calculated using your method was 59.09073. I'm guessing this was due to rounding. Oct 23 '20 at 13:42
• Good catch. Thanks for the nice problem. Oct 23 '20 at 14:08

Searching for "right triangle math" brings back fun memories from high school. There are dozens of online calculators that will let you plug in the values and give you the answer, but I think you want to know the backbone of the process.

The Math Is Fun web site has a relatively clear and uncluttered explanation.

You have to know a couple of pieces of the puzzle but your drawing contains all of that. Even missing a bit of information, you can figure out the segment to calculate. The rest of the site presents the use of sin, cos and tan for creating the formulae necessary to get the answer.

In my high school days, it was a breeze to memorize the stuff, but I fall back to the calculators online nowadays.

• This does not answer the question as this is not a simple triangle with sides of 0 width. What hypotenuse to use? What is the length of the adjacent to use? Oct 20 '20 at 18:42
• But it is still basic Trig Mascaro. And Fusion should be able to tell you directly. Search Fusion measure distance. Oct 20 '20 at 18:47
• It's not basic trig. If it was basic trig, why can you not post a correct answer? Oct 21 '20 at 18:10