<s>[INST] Area of a triangle ABC Triangle's $ABC$ angle $C=60°$, $AB = AC+2 = BC-1$.
Find the area of this triangle.
I've tried writing $AB$ as $x$ so $AC= x-2$ and $BC = x+1$. 
Then i calculated the area with 
$$\frac12\sin(60°)\cdot(x-2)\cdot(x+1)=\frac12\sin(60°)\cdot(x^2-x-2)$$
Now i have no idea what to do next. 
 [/INST] With the theorem of cosines we get
$$c^2=a^2+b^2-2ab\cos(\frac{\pi}{3})$$
substituting $$c=b+2,a=b+3$$ we get an equation for $b$:
$$(b+2)^2=(b+3)^2+b^2-2(b+3)b\cos(\frac{\pi}{3})$$
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