# Finding side "x" (triangles)

• Apr 3rd 2010, 06:47 PM
shawli
Finding side "x" (triangles)
Can't figure this one out.

The answer must be in exact value.

*Picture has been fixed to add the missing side length
• Apr 3rd 2010, 07:01 PM
Sudharaka
Quote:

Originally Posted by shawli
Can't figure this one out.

The answer must be in exact value.

Dear shawli,

I think to solve this problem the lengh of at least one of the sides must be given. Are you sure the length of any side is not given??
• Apr 3rd 2010, 07:02 PM
Prove It
Quote:

Originally Posted by shawli
Can't figure this one out.

The answer must be in exact value.

There's not enough information. You need to give at least one known length.
• Apr 3rd 2010, 07:07 PM
shawli
Sorry! I missed a label, but I've reposted the picture with the correction.
• Apr 3rd 2010, 07:11 PM
Sudharaka
Quote:

Originally Posted by shawli
Sorry! I missed a label, but I've reposted the picture with the correction.

Dear shawli,

If the other side of the rectangle is y;

$\displaystyle tan45^0=\frac{x}{y}$

$\displaystyle tan60^0=\frac{x+4}{y}$

Can you do it from here??
• Apr 3rd 2010, 07:14 PM
Prove It
You're expected to be able to recognise triangles that are of dimensions $\displaystyle (1, 1, \sqrt{2})$ and $\displaystyle (1, \sqrt{3}, 2)$.

Looking at the large triangle, its dimensions are

$\displaystyle (x, \sqrt{3}\,x, 2x)$.

So the base of this shape is $\displaystyle \sqrt{3}\,x$.

Using that information, you should be able to see that

$\displaystyle \sqrt{3}\,x - x = 4$

$\displaystyle x(\sqrt{3} - 1) = 4$

$\displaystyle x = \frac{4}{\sqrt{3} - 1}$

$\displaystyle x = \frac{4(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$

$\displaystyle x = \frac{4(\sqrt{3} + 1)}{3 - 1}$

$\displaystyle x = \frac{4(\sqrt{3} + 1)}{2}$

$\displaystyle x = 2(\sqrt{3} + 1)$

$\displaystyle x = 2\sqrt{3} + 2$.