# Determining angle to cut two boards

I am building a small roof and I need to determine how to find the distance/angle in that I need to cut two 2x4's in order to make them fit together perfectly. I've drawn a crude image below to illustrate.

Given that my base is 6', and I'd like to have the center of the roof 2' high from the base, what formula can I use to determine how long a and b will be as well as the angle that boards a and b will come together?

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Quite possibly the single most applied question I've ever seen on this site! –  Steven Stadnicki Mar 26 '12 at 16:31
mycarpentry.com/roof-framing.html –  user996522 Mar 26 '12 at 16:33
Very nice question, clearly put. –  André Nicolas Mar 26 '12 at 17:06

If the base is $2c$ and the height is $h$, then the roof framing cross sections are shown in blue below. Note that the triangles with sides $ach$ & $bch$ are congruent (mirror images, reflections about $h$), so really $a=b$ have the same length. (We only confuse things by using two variable names for them!) Furthermore, this length is the hypotenuse of a right triangle with horizontal and vertical legs $c$ & $h$, respectively, and we can use its trigonometry to find its complementary interior acute angles, which I will call $\alpha$ and $\beta$.
Now the angle $\beta$ opposite $h$, the roof's angle, has slope (or pitch) $\frac23$, from which we get $\beta$ using the arctangent (inverse tangent) function (with your calculator's angle mode set to degrees rather than radians): $$\tan\beta=\frac{h}{c}=\frac23\qquad\implies\qquad\beta =\tan^{-1}\frac23=0.588\text{ rad}=33.69^\circ$$ and the angle you want (opposite $c$ inside each congruent triangle at the top) is $$\tan\alpha=\frac{c}{h}=\frac32\qquad\implies\qquad\alpha =\tan^{-1}\frac32=0.9828\text{ rad}=56.31^\circ.$$ That is, you need to cut an acute angle $\alpha=90^\circ-\beta$ on each piece of wood, to get a total angle of $2\alpha$ at the top of the roof. As to the lengths, they are given by the famous Pythagorean formula: \eqalign{ a^2=b^2&=h^2+c^2\\&=2^2+3^2\\&=4+9\\&=13 \\\\ a=b &=\sqrt{13} \approx 3.60555\text{ ft} \\ &\approx 3\text{ ft }7.26661\text{ in} \\ &\approx 3\text{ ft }7\tfrac4{15}\text{ in} \\ &\approx 3\text{ ft }7\tfrac14\text{ in} \\ &\approx 109.9\text{ cm} } The alternate length unit and fraction are slightly more precise, but perhaps less convenient to work with.
Assuming you want $a=b$, you can draw a right-angled triangle sides $3$ and $2$ and hypotenuse $a$. So $a = \sqrt{(3^2 + 2^2)}$. Then the angle between $a$ and the horizontal is $\arctan(2/3)$, and the angle between a and b is $2 \arctan(3/2)$