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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.

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  • $\begingroup$ If $AB=x$ then $AC=x-2$, $BC=x+1$ $\endgroup$ – Vasya Oct 2 '19 at 18:45
  • $\begingroup$ Well, immediately next is replace $\sin 60 $ with $\frac {\sqrt 3}2$.... then after that...;) Well, then after that, solve for $x$ with either/or law of sines or cosigns. $\endgroup$ – fleablood Oct 2 '19 at 19:00
  • $\begingroup$ Law of cosines gives us $AB^2 = AC^2 + BC^2 - 2AC*BC\cos 60$ and as $\cos 60 =\frac 12$ this is a quadratic equation that will give us $x$. $\endgroup$ – fleablood Oct 2 '19 at 19:02
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If $AB=x$, then $AC=x-2,BC=x+1$.

Now use the cosine formula: $$x^2=(x-2)^2+(x+1)^2-(x-2)(x+1)$$ Expanding we get $x=7$. So the longest side is 8 and the height $AC\sin60^o=5\sqrt3/2$. Hence the area is $10\sqrt3$.

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Let $AC=x$ then $AB=x+2$ and $BC=x+3$.

To find $x$ we can use

Finally we can evaluate the area with the formula you have already mentioned.

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  • $\begingroup$ You should probably proofread this for errors. $\endgroup$ – Ted Shifrin Oct 2 '19 at 18:54
  • $\begingroup$ @TedShifrin Opsss...thanks I fix! $\endgroup$ – user Oct 2 '19 at 18:55
  • $\begingroup$ What is $A$ in your first equation? $\endgroup$ – Ted Shifrin Oct 2 '19 at 18:57
  • $\begingroup$ It certainly wasn't clear to me when $A$ is appearing six or seven other times as a vertex. $\endgroup$ – Ted Shifrin Oct 2 '19 at 18:58
  • $\begingroup$ @gimusi: $BC=x+3$, no? $\endgroup$ – Vasya Oct 2 '19 at 18:59
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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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