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


Now i have no idea what to do next.

  • $\begingroup$ 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

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

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

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.