# Finding side on triangle

• Jun 22nd 2010, 05:41 AM
nerdzor
Finding side on triangle
Find x
• Jun 22nd 2010, 05:43 AM
Ackbeet
6 what? Degrees? 15 degrees? 200 what? Meters? Maybe they didn't give you any units. That's ok.

What have you thought of doing so far?
• Jun 22nd 2010, 05:44 AM
nerdzor
6 and 15 are degrees and the 200 has no units.

I've found all angles on the triange and don't know where to go from there
• Jun 22nd 2010, 05:47 AM
Ackbeet
I would slap a label on the segment of the horizontal just to the right of the 200. What do you want to call that?

Further question: are we right in assuming the lower-right angle is a right angle?
• Jun 22nd 2010, 05:48 AM
nerdzor
Yes. The lower right is a right angle. I named the segment to the right of 200 'y' and the hypotenuse 'z'
• Jun 22nd 2010, 05:50 AM
Ackbeet
Great. Now, just thinking out loud here. You could use the Pythagorean theorem, since you have a right triangle. The only problem with that is that it introduces yet another variable for which you'd need another equation. What's a relevant, correct equation you could write down that involves x?
• Jun 22nd 2010, 05:52 AM
nerdzor
I tried sin cos and tan for the bottom left hand corner angle (6 degrees) and the angle at the top (84 degrees) and tried to equate some of the equations that involved x but hasn't worked out so far.
• Jun 22nd 2010, 05:54 AM
Ackbeet
Ok, well, show me what you have so far. We'll go from there.
• Jun 22nd 2010, 06:01 AM
nerdzor
$sin 6 = \frac{x}{2}$

$cos 6 = \frac{200+y}{z}$

$tan 6 = \frac{x}{200+y}$

$sin 84 = \frac{200+y}{z}$

$cos 84 = \frac{x}{z}$

$tan 84 = \frac{200+y}{x}$
• Jun 22nd 2010, 06:04 AM
Ackbeet
Of those equations, I would agree with all but the first one.

You're going to need some equations having to do with the inner triangle, or I'm mistaken. What can you get from the inner triangle?
• Jun 22nd 2010, 06:06 AM
Ackbeet
Oh, and a LaTeX hint or two: use \cos instead of cos. Also, I find it a good idea always to put parentheses around the argument. Thus, $\sin(6^{\circ})$, not $\sin 6^{\circ}$. The reason is that sometimes, the arguments get to be pretty hairy, and you can get some ambiguous expressions. It's better to be safe!