Geometry...

[Nineshadow]

 

We've got a cube, ABCDA'B'C'D'.

{O}= BC' ∩ CB'

 

We need to find :

sin(CD' , AO)                          (the sine of the angle between CD' and AO)

 

Can you do it?

It took me some time but I finished it.

[Foxxer]

I don't know how did I passed geometry classes in high school.

[Nineshadow]

I don't know how did I passed geometry classes in high school.

I'm not even in high school...

8th grade.

[Osmium]

I'm not even in high school...

8th grade.

 

I could do it if I had a calculator but my sister stole mine so I can't. That shouldn't be too difficult considering it is a cube though.

 

The way I'd do it is plug in a side length for the lwh and use pythagorean theorum to find the diagonals of the squares, then divide that by two to have half the distance of the diagonal, then pythagorean theorum to get the hypotenuse and then I have all three side lengths.

[Ciccioo]

but CD' and AO are skew

where is the angle?

[Nineshadow]

I could do it if I had a calculator but my sister stole mine so I can't. That shouldn't be too difficult considering it is a cube though.

Calculators are bad for Math problems that have real numbers.

sqrt 2 isn't equal to 1.41.It's almost 1.41.

 

That's why you have to do it without a calculator which aproximates those numbers.

 

but CD' and AO are skew

where is the angle?

There is.You can draw parallels to one of them or both of them.

[Osmium]

Calculators are bad for Math problems that have real numbers.

sqrt 2 isn't equal to 1.41.It's almost 1.41.

 

That's why you have to do it without a calculator which aproximates those numbers.

 

That is true but it takes more than this to get me to want to do it on paper

[The Crazed Child]

This is why I love the LTT forums...you can ask anything you want from help getting friends to geometry problems.