Angle calculation

[Claretdcwat]

Hello,

I’m sure that this is probably a really simple question but I’m struggling to remember back to my o level mathematics days!

If I have an object that measures 379mm in depth and a space of 350mm to place it on, what angle do I need to raise the object by for it to fit within the 350mm space (height no object but cannot be vertical).

Effectively, I’m wanting to fit a box in a until where the box is deeper than the unit allows.

What is the formula / calculation that I should use?

Thank you for any help.

 

[Dr.Peterson]

Hello,

I’m sure that this is probably a really simple question but I’m struggling to remember back to my o level mathematics days!

If I have an object that measures 379mm in depth and a space of 350mm to place it on, what angle do I need to raise the object by for it to fit within the 350mm space (height no object but cannot be vertical).

Effectively, I’m wanting to fit a box in a until where the box is deeper than the unit allows.

What is the formula / calculation that I should use?

Thank you for any help.

Please attach a picture of the situation, so we can be sure of what you want to do. What shape are the "object" and the "place"? What does "fit within" mean?

 

[The Highlander]

Hello,

I’m sure that this is probably a really simple question but I’m struggling to remember back to my o level mathematics days!

If I have an object that measures 379mm in depth and a space of 350mm to place it on, what angle do I need to raise the object by for it to fit within the 350mm space (height no object but cannot be vertical).

Effectively, I’m wanting to fit a box in a until where the box is deeper than the unit allows.

What is the formula / calculation that I should use?

Here is a complete guess at your situation.
(See attached diagram.)

If you are to fit a 379 mm long line into a 350 mm 'space' (we prefer to talk about height rather than depth as that can be an ambiguous term) then the "calculation" is, indeed, "simple": ?

\(\displaystyle sin θ=\left(\frac{350}{379}\right) \Rightarrow θ=sin^{-1}\left(\frac{350}{379}\right) ≈ 67.44°\)
(and the other angle in my sketch is just 180 - θ).​

However, as I trust you can see from the sketch, a line has no width! Any "object" ("box" or otherwise) must have height, width & depth (distance from front to back when "facing" the object; you see how that's such an ambiguous term when quoted on its own?). ?
As soon as we include any second dimension to the object in the sketch (qv) then it becomes a whole different calculation and, of course, your "box" will have not two but three dimensions to deal with. ?
You have, perhaps, been thinking that at least one of these might be 'discounted' (in the situation you, somewhat vaguely, describe) but can you now see how, even without a sketch of your problem, you haven't provided anywhere near enough (dimensional) information for us to offer any "help". ?

If you want (really need?) further "help" please provide sketch(es), drawing(s), even pictures (too) might be helpful, with all the dimensions you can measure shown.
Though, to be honest, (given my limited understanding of your situation as you describe it) I am at a loss to understand why you would even care what the "angle" is. Just tilt the "box" until it fits into the "unit"; why worry about what the angle it's sitting at is? ?