# Easy Trigonometry Problem

• Mar 18th 2009, 10:42 AM
Mr Rayon
Easy Trigonometry Problem
The angle formed by the diagonal of a rectangle and one of its shorter sides is 60°. If the diagonal is 8 cm long, find the dimensions of the rectangle, in surd form.

I am having trouble imagning this. I am also not sure what the question is asking me to do.

Any help regarding the problem above will be appreciated!
• Mar 18th 2009, 10:48 AM
Moo
Hello,

Attachment 10570

Just apply the definition of cosine and sine in the right-angled triangle ABC :

$\displaystyle \cos 60=\frac{AC}{BC}$

$\displaystyle \sin 60=\frac{AB}{BC}$

and for the values of cos and sin, they're just common values.
• Mar 18th 2009, 10:53 AM
Mr Rayon
Quote:

Originally Posted by Moo
Hello,

Attachment 10570

Just apply the definition of cosine and sine in the right-angled triangle ABC :

$\displaystyle \cos 60=\frac{AC}{BC}$

$\displaystyle \sin 60=\frac{AB}{BC}$

and for the values of cos and sin, they're just common values.

Ok...thanks for the diagram. Now I know what I'm looking at! But I'm afraid I'm still a little lost. (Crying)
• Mar 18th 2009, 11:00 AM
Mr Rayon
Nevermind...I get it now!

EDIT:

See:

cos 60 = AC/BC
1/2 = x/8
8/2= x
x= 4

(√3)/2 = AB/BC
(√3)/2 = x/8
(8√3)/2 = x
x= 4√3
• Mar 18th 2009, 11:20 AM
Moo
Quote:

Originally Posted by Mr Rayon
Nevermind...I get it now!

EDIT:

See:

cos 60 = AC/BC
1/2 = x/8
8/2= x
x= 4

(√3)/2 = AB/BC
(√3)/2 = x/8
(8√3)/2 = x
x= 4√3

Yes (Nod)
• Jun 17th 2010, 11:11 PM
Mr Rayon
Quote:

Originally Posted by Moo
Hello,

Attachment 10570

Just apply the definition of cosine and sine in the right-angled triangle ABC :

$\displaystyle \cos 60=\frac{AC}{BC}$

$\displaystyle \sin 60=\frac{AB}{BC}$

and for the values of cos and sin, they're just common values.

Hey, moo how'd you draw the right angle triangle? What software/file do you use? How can I do the same thing in a post in this forum?
• Jun 30th 2010, 08:40 AM
Moo
Drew it in paint, then used the attachment option in a post to add it to my post.