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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 $x$?

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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=$2x=10\sqrt 3$

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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
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Hint: Use the law of cosines.

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