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This is an illustration of two brackets. I want to build a trapezoidal shelf on it. I currently have sufficient 11 and one quarter boards to work with for whatever the answer is for the problem I'm about to give you. The distance between the two brackets is 16 inches. One bracket is 8 inches, the longer one is 18 inches. You might be able to see the smaller bracket has two holes and the larger one has four with which to anchor the two boards built by appropriately cutting the 11 and a quarter inch boards that I have. What I need from you are dimensions and angles to make it so that The two boards are cut appropriately to allow an equal number of holes to anchor each board so that the final shape is a trapezoid and so that each board maintains a proportionality that is smooth. That is, at the small bracket, each piece should be along the line parallel with the bracket four inches at the point where it rests on the large bracket because it's 18 inches long, each board should be nine inches. Have I sufficiently described the problem for you?
G: [Non-text content: thoughts; content_type=thoughts]
G: [Non-text content: reasoning_recap; content_type=reasoning_recap]
G: Yes — you described it enough. Interpreting your setup as a plan-view trapezoid:
That’s clean and proportional, and it lets you allocate holes evenly (small bracket: 1 screw per board; large bracket: 2 screws per board).
Let the wall be the 16" back edge.
Total shelf outline
Center seam line (where the two boards meet)
This one is easiest.
Make a rectangle blank 16" long and 9" deep (9" is the max depth this piece ever needs).
Mark on the two ends:
Connect those depth marks and cut the diagonal.
So Board A corners (in inches), using wall as the baseline:
That diagonal cut (seam edge) has an angle to the wall of: [ \arctan\left(\frac{9-4}{16}\right)=\arctan(0.3125)=17.35^\circ ]
Same end depths as Board A, but its two long edges are different lines.
This piece is bounded by:
The front edge angle to the wall is: [ \arctan\left(\frac{18-8}{16}\right)=\arctan(0.625)=32.00^\circ ]
Practical way to lay out Board B without fighting angles:
That template method is the “no surprises” approach and avoids compound-thinking entirely.
That matches your “smooth proportionality” goal and keeps the load shared.
If you want, tell me the spacing of the holes on the 18" bracket (distance from wall to each hole center), and I’ll tell you exactly where the seam should fall relative to them so each board lands cleanly on “its” two holes (instead of the seam landing awkwardly right through a hole line).
M: What is the angle for the long end, the long edge, of a trapezoid whose other dimensions are four inches, 16 inches, and 18 inches? Sorry, I phrased that badly. What is the angle between the four-inch side and the irrational long side of that trapezoid? I'm using a circular saw and I can change the angle, but I need to know what it would be to. Don't worry about trying to give me the appropriate angle here. I can do the basic math, but I need something to work with.
G: [Non-text content: thoughts; content_type=thoughts]
G: [Non-text content: reasoning_recap; content_type=reasoning_recap]
G: Assuming the trapezoid is the “wall edge” 16" long, with perpendicular end depths of 4" (short end) and 18" (long end), then the “irrational long side” is the slanted front edge connecting the 4" tip to the 18" tip.
Angle between the 4" side (the perpendicular end) and the slanted long side:
That’s the complement of the slope angle:
[ \theta = \arctan\left(\frac{16}{14}\right) = 90^\circ - \arctan\left(\frac{14}{16}\right) \approx 48.815^\circ ]
So: ≈ 48.8°.
Angle between the slanted long edge and the 16" edge (the wall-parallel edge):
[ \phi=\arctan\left(\frac{14}{16}\right)\approx 41.185^\circ ]
So: ≈ 41.2°.
If you tell me whether your saw’s scale expects the angle off-square (miter relative to a 90° crosscut) or the angle off the board edge (relative to 0°), I can map that 48.8° into the exact “set it to this” number without ambiguity.
M: I knew I screwed that up. No, I'm sorry. Not an 18-inch edge, a 9-inch edge. I forgot to have both sides because the transcriber might screw that up. I mean that I forgot to cut the value of both brackets in half before asking you that question. So it's a 4-inch by 16-inch by 9-inch trapezoid. I'm trying to figure out the angle there at that same point.
G: Good — that fixes the geometry.
Now we have:
[ 9 - 4 = 5" ]
[ \arctan\left(\frac{5}{16}\right) = 17.35^\circ ]
That is the angle the long slanted edge makes relative to the 16" side.
Angle between the 4" side (vertical) and the slanted long side
That angle is:
[ 90^\circ - 17.35^\circ = 72.65^\circ ]
≈ 72.65°
Most saw miter scales read angle off square (90°).
If you're cutting from a square end and rotating the saw to create the slope:
Set the saw to:
17.35°
That produces the correct trapezoid angle.
If your saw instead reads angle relative to straight (0°), you'd use:
72.65°
But 99% of circular/miter saws use the 17.35° setting.
If you want, I can also give you the slanted edge length so you can double-check your cut with a tape before installing.