I’m working on a really large and complex model and I’ve come across with what, at first, may seem a easy problem.

Well, I’m trying to draw a right triangle from the hypotenuse and one side, but I know no angles (except the 90º one, of course).

I’ve tried to draw a circle with the known side as the radius, with the center on one end of the hypotenuse and then draw a line from the other end tangent to the circle, but this method is slow and inaccurate.
Then I drew a random right triangle and by using pythagoras theorem scale it on the second side axis, but I’d have to do this 73 times in the model and I’d go crazy…

Do you know any plugin or easier way to solve this?

Perhaps it would be quickest to solve the unknown lengths with your favorite spreadsheet application.
Simple to draw given all sides and they’re all right triangles.

There’s this SketchUp plugin, but it requires 3 known sides. Trilateration - v1.1 by Didier Bur
Command to draw a triangle, knowing its 3 side lengths (trilateration)

There are some plugins that will draw a line perpendicular to a given line - you could try looking them up.

Here’s how I would do it using built-in tools (this is actually easier than it might sound at first, though admittedly doing it for 73 triangles might get tedious):

Select the protractor tool

Click and drag a little way along the line (this will orient the protractor perpendicular to the line)

Click on the far end of the hypotenuse. This will set the reference angle of the protractor.

Move the cursor a bit until you see a value in the VCB. Then enter 0 in the VCB. This will draw a construction line perpendicular to the line and in the plane of the triangle.

Use the tape measure tool to create another construction line parallel to the one in step 4 but through the far end of the hypotenuse (that is, click and drag the CL over to the end of the hypotenuse).

Use the line tool to fill in the third side of the right triangle along the CL from step 5.

PS - Shep’s method is the same except it relies on the lines being in a cardinal plane so that the protractor naturally selects the same plane as the triangle. My steps 2-4 work for any orientation.

If you just need the shape as a guide, how hard would it be to make a dynamic component of 3 lines where you just input 2 sides and it calculates the rest and moves them for you.
(I’m sure there’s plugins, I just have a habit of using native tools first, unless its a big time saver, this seems like it won’t take too long)

Thanks pointing that out. And, it’s even easier than that. In your step 4, you don’t need to enter 0, simply click on the far end of the hypotenuse a second time.

I have never considered or been in the need of using them, maybe the time has arrived
I think I am trying that too, let’s see what I can get. Thank you very much!

Okay, I’m all in with this. Grab the stick by the other end.

Draw a line 90° from the end of the known side a bit longer than the hypotenuse.
Grab the “pie” arc tool a place it at the original angle.
Start the arc at the far end of the hypotenuse and swing it around to the line drawn earlier till you see the intersection inference.
Draw a line from the arc/line intersection back to the original angle.
Erase the arc, select the right triangle and rotate back to the starting angle if desired.

Aha, we misread your original post. You don’t have a second line, you just have a length.

Shep’s second proposal works using either the “arc” or “pie” tools because these tools place their final point at exactly the same radius as their start point. You don’t actually care about the arc between them, so unlike your method of trying to construct a tangent, it doesn’t matter that the arc is only approximated by a collection of line segments. The radius is exact.

The older version of the Arc or Circle tool didn’t catch this inference. I think the Pie tool can be used to solve several geometry issues that the older tools couldn’t do.

I probably showed up too late for this to really matter.

But I’d like to add in one more possible approach, even if it’s only useful to someone who’s reading through old threads at some future time.

The point I want to make is that. . .

If you use the hypotenuse as the diameter of a circle. And thus draw a circle which has it’s center point based upon the mid point of the hypotenuse (with a radius set to both ends of the hypotenuse line).

From that setup, angles can be drawn in any direction from the end points of the hypotenuse line – and a 90º angle will result at any intersection point that’s made on the circles circumference between the two lines.

Probably easier seen in the attachment, than to picture based on my description above.