Rhombus

[minko]

I got question that one side of rhombus is long 6cm and thats a and alpha is 44*
so i needed to calculate how long are diagonals e and f
Did i get correct answers ?

 

[HallsofIvy]

I would be inclined to use the "cosine law": if a triangle has side a, b, c and the angle between sides a and b is C, then \(\displaystyle c^2= a^2+ b^2- 2ab cos(C)\).

Here you are given that all four sides have length a= 6 cm and that one angle is \(\displaystyle \alpha= 44\) degrees. You have correctly determined that the other angle is \(\displaystyle \beta= 180- 44= 136\) degrees.

The longer diagonal, opposite the angles \(\displaystyle \beta\), we have \(\displaystyle d_1^2= 2(6^2)- 2(6^2)cos(136)\).

For the shorter diagonal, opposite the angles \(\displaystyle \alpha\), we have \(\displaystyle d_2^2= 2(6^2)- 2(6^2)cos(44)\).

 

[minko]

I would be inclined to use the "cosine law": if a triangle has side a, b, c and the angle between sides a and b is C, then \(\displaystyle c^2= a^2+ b^2- 2ab cos(C)\).

Here you are given that all four sides have length a= 6 cm and that one angle is \(\displaystyle \alpha= 44\) degrees. You have correctly determined that the other angle is \(\displaystyle \beta= 180- 44= 136\) degrees.

The longer diagonal, opposite the angles \(\displaystyle \beta\), we have \(\displaystyle d_1^2= 2(6^2)- 2(6^2)cos(136)\).

For the shorter diagonal, opposite the angles \(\displaystyle \alpha\), we have \(\displaystyle d_2^2= 2(6^2)- 2(6^2)cos(44)\).

Can we use pythagoras theorem to get e and f sides ?

 

[HallsofIvy]

I don't know what you mean by "e and f sides". Do you mean the two diagonals? I suppose you could but why? You would just end up re-deriving the cosine law.

 

[minko]

I don't know what you mean by "e and f sides". Do you mean the two diagonals? I suppose you could but why? You would just end up re-deriving the cosine law.

Yes i meant diagonals, i thought that law of sines we can use only in triangle.

 

[Cubist]

On the RHS of your answer you seem to have used tried using the cosine law, but you have written "sin" instead of "cos".
For this reason your value of "f" is wrong.

The LHS of your answer obtains "e" very quickly. Why not use a similar method to obtain "f"?

 

[minko]

On the RHS of your answer you seem to have used tried using the cosine law, but you have written "sin" instead of "cos".
For this reason your value of "f" is wrong.

The LHS of your answer obtains "e" very quickly. Why not use a similar method to obtain "f"?

So for the diagonal e is right answer ?

 

[Cubist]

On a general note, be very careful when you write

[math]\frac{f^2}{2}[/math]
when you actually mean

[math]\left( \frac{f}{2} \right)^2[/math]
these two have quite different values!

 

[Cubist]

So for the diagonal e is right answer ?

Yes, approximately. sin(68°)*6*2 ≈ 11.1262

 

[minko]

Yes, approximately. sin(68°)*6*2 ≈ 11.1262

An for the diagonal f i got answer 4.5 cm and f/2 is 2.256cm, i think thats correct

 

[Cubist]

An for the diagonal f i got answer 4.5 cm and f/2 is 2.256cm, i think thats correct

That's approximately correct. Another way of working it out would have been

f = cos(68°)*6*2 ≈ 4.4953cm

OR Pythagoras

[math] \left(\frac{f}{2}\right)^2 = 6^2- \left(\frac{e}{2}\right)^2 [/math]

 

[minko]

That's approximately correct. Another way of working it out would have been

f = cos(68°)*6*2 ≈ 4.4953cm

OR pythagoras

[math] \left(\frac{f}{2}\right)^2 = 6^2- \left(\frac{e}{2}\right)^2 [/math]

Thank you very much for your help