# Cosine law

1. Aug 20, 2008

### sky08

hello , How are you all?
look at this question
what the magnitude of the angle that between two ribs from the heads of the cube to its center ?
I attempted by using cosine law [look at the pecture in attachment]
but, Hwo to apply this law on the cube (3D)?
any one help me >>and thank you anyway

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2. Aug 20, 2008

### HallsofIvy

Staff Emeritus
I am afraid that I don't understand what you mean by "ribs" or "heads" of a cube. Are the "ribs" diagonals and the "heads" vertices? In that case, the angle between two diagonals is a right angle. You don't need the cosine law to see that.

3. Aug 20, 2008

### sky08

yes ,,
look at attachment the picture is clear that explanation
I calculat it by using simple way but How to use cosine law to giveangle theta in pecture
an thanks Mr.HallsofIvy

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• ###### part of a cube.png
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4. Aug 20, 2008

### snipez90

Hmm in the drawing is theta the same as angle BOA? Well either way it's BOA or half of BOA due to symmetry so you can find BOA easily and divide if you have to.

Angle BOA has two $$\frac{\sqrt{3}r}{2}$$ sides that make up Angle BOA. The law of cosines tells us that $$(AB)^2 = (OA)^2 + (OB)^2 + 2(OA)(OB)cos(Angle BOA)$$. Then just solve for Angle BOA.

Although really you shouldn't need the law of cosines.

5. Aug 20, 2008

### sky08

Hmm in the drawing is theta the same as angle BOA? Well either way it's BOA or half of BOA due to symmetry so you can find BOA easily and divide if you have to.

sorry , my explanation not clearly anyway

now I understand

used cosine law directly and I don't need analysis all angles because I can find it from symmetry considerations .
thank you very much

6. Aug 20, 2008

### BoundByAxioms

Err, the formula is $$(AB)^2 = (OA)^2 + (OB)^2 - 2(OA)(OB)cos(Angle BOA)$$.

7. Oct 27, 2008

### sky08

thanks Mr. BoundByAxioms for your correcting :>
I hope the best for all