Area of a triangle

That sure is a very nice problem you have there. But what are your thoughts? What have you tried? Did you draw a picture? Use any kind of trig functions?

Please (re-)read the Read Before Posting thread that's stickied at the top of each sub-forum and comply with the rules found therewithin. Specifically, please share with us any and all work you've done on the problem, even the parts you know for sure are wrong. Thank you.
 
I've tried writing AB as x so AC= x-2 and BC = x+1. Then i calculated the area with 1/2*sin(60°)(x-2)(x+1) = 1/2*sin(60°)*(x^2-x-2). Now i have no idea what to do next.
 
What you've done so far is good. That gives you the area of the triangle in terms of \(x = \overline{AB}\). So now you need to figure out a way to determine the value of \(x\). This is why I asked if you'd drawn a picture. It can be very helpful to figure out what to do.

triangle.jpg

You only know one angle so that rules out the law of sines... what about the law of cosines? Plugging all the information into the formula:

\(\displaystyle x^2 = (x+1)^2 + (x-2)^2 - 2(x+1)(x-2) \cos(60^{\circ}) \)

Where does that lead you?
 
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