You are right, and the angle will be 70,21562478 
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It will take time to get that level of experience in inferencing, though.
Guides provide insight, and can take care of more complex shapes, be faster, sometimes etc
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are these shapes all the same? How about a theory that the extended corner lines must create a true pyramid with the apex directly above the center? Then you have a reference for rotation of the already drawn faces. Oh but this is like a problem we had before. Without a plugin. SU canât make that reference accurately.
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The sides are the same, and there is indeed a formula for determining the angle of a four-sided pyramid! But I had hoped thereâd be a formula that would give me the angle without determining the height of the triangle. Such is life!
SU canât make that reference accurately.
I didnât know Sketchup had that sort of inaccuracy?
Itâs just that thereâs no way to say:-take this point and rotate until it hits the line. You can snap to the line but it can be anywhere along the line (or plane). Also, though there are plugins that do calculate accurately, you cannot use an arc to find the point since arcs in SU are not true arcs.
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Sorry but I donât buy that (yet). Show me all the dimensions of the four faces in your original face so I can see if SketchUp canât do what you are asking.
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The only limitation I know of is that SketchUpâs units format will display at most three decimal places even if the internal value is more precise. But it is pretty rare that 0.001 degree isnât good enough!
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Sorry @MBGM-MBGM , I thought you made that remark about SketchUp not being accurate. I must be paying more attention when reading, seeing that in fact it was @pbacotâŚ
Yeah, itâs not the accuracy of what SU draws. Itâs the question of some way to make an inference to rotate these shape exactly where they need to be without some other trigonometric calculating. We know two sides of the right triangle so we should be able to find the pointâjust how could you do it with SU drawing tools alone? There was another puzzle a long while ago, that dealt with the inability of using SU arcs to solve an accurate rotation. Actually that puzzle might have been solved.
Fredo6 dealt with that issue in the Free Rotate tool in FredoScale but that was before the release of the Pie and Arc tools.Now itâs possible with the native tools as has been shown.
OK. I hope so.
using Pythagoras theorem i worked out the following
- give the angled edge in MBGM-MBGMâs face the name H, for Hypotenuse.
- give the height of the face the name h, for height.
- give the base of the triangle made by H and h the name b, for base.
- note that in plan view the leaned in faces will touch along the 45degree angled lines passing through the corners of the âcorralâ. This means that if you drop a vertical from where the Hâs touch in the final desired shape it will touch the ground exactly distance b from the bottom of MBGM-MBGMâs face. Give the dropped vertical the name l, for lengthUnkown.
- By Pythagoras, the length of the diagonal linking the bottom corner of the desired shape to this point is the square root of two times b squared. Give this diagonal line the name d, for diagonal.
- note that l squared plus d squared equals H squared, again by Pythagoras.
- note that d squared equals the square root of two times b squared, from 5 above.
- so l squared plus two times b squared equals H squared therefore l (lengthUnknown) equals the square root of H squared (known) minus twice b (known) squared. l is now known.
You can easily set up the triangle along the diagonal and rotate MBGM-MBGMâs face to the top of l.
The results of your post can be written in the compact trigonometry form as âcos(phi) = 1/tan(theta)â where theta is 71.3 deg (specified angle of side) and phi is the angle through which the side must be rotated in order to make the corners meet. Your calculator likely has the inverse function for cos. See the attachment for a derivation.angle sided box.pdf (157.5 KB)