# 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.

• If $AB=x$ then $AC=x-2$, $BC=x+1$ – Vasya Oct 2 '19 at 18:45
• 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. – fleablood Oct 2 '19 at 19:00
• 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 $x$. – fleablood Oct 2 '19 at 19:02

## 3 Answers

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 $$10\sqrt3$$.

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

To find $$x$$ we can use

• $$AB^2=AC^2+BC^2-2\cdot AC\cdot BC \cos (60°)$$ (law of cosine)

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

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

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 $$b$$: $$(b+2)^2=(b+3)^2+b^2-2(b+3)b\cos(\frac{\pi}{3})$$