Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've a isosceles triangle which length is $10\;\mathrm{cm}$ , $10\;\mathrm{cm}$ and $x$. If I want to make this triangle $120^\circ$ degree then what should be the $x$?

share|cite|improve this question
up vote 2 down vote accepted

Angles opposite to equal sides are equal, so one angle is $120^0$ while others are $30^0$ each (angles opposite to equal sides can't be $120^0$,otherwise sum of angles of triangle would be greater than $180^0$). Draw perpendicular from vertex (intersection of two equal sides) to the opposite side, it divides the opposite side into two equal halves. Let one half of that be $x$,then $$\frac{x}{10}=\cos30^0\implies x=5\sqrt 3$$ Thus the side length=$2x=10\sqrt 3$

share|cite|improve this answer
But also first rule out the possibility that the two equal angles are $120^\circ$. – GEdgar Jul 12 '12 at 19:24
Then the third angle is minus 60 degrees (180 - 120 - 120). – marty cohen Jul 13 '12 at 4:24
i edited my answer and mentioned the possibility.check it. – Aang Jul 13 '12 at 4:56

Hint: Use the law of cosines.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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