Take a point \(O\) from which three line segments \(OA, OB, OC\), lying on the same plane, emanate such that their respective lengths are \(a, b, c\) and the angle between any two of them is 120 degrees. Use Cosine rule and the mentioned equations to find \(AB, BC, CA\) which will have values 3,4 and 5 units. Now this triangle \(ABC\) is right angled( by the converse of Pythagoras' Theorem), so it's area will be 6 sq. units. Also, the area of the three triangles \(AOB, BOC, COA\) can be found out by the sine rule (individually), and now equate it with the earlier found area. The answer comes out to be \(8 \sqrt{3}\). [ the calculations are to be done by you].

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TopNewestTake a point \(O\) from which three line segments \(OA, OB, OC\), lying on the same plane, emanate such that their respective lengths are \(a, b, c\) and the angle between any two of them is 120 degrees. Use Cosine rule and the mentioned equations to find \(AB, BC, CA\) which will have values 3,4 and 5 units. Now this triangle \(ABC\) is right angled( by the converse of Pythagoras' Theorem), so it's area will be 6 sq. units. Also, the area of the three triangles \(AOB, BOC, COA\) can be found out by the sine rule (individually), and now equate it with the earlier found area. The answer comes out to be \(8 \sqrt{3}\). [ the calculations are to be done by you].

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Is it possible to solve this algebraically?

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Well, your idea is interesting, but are you sure about the value of ab + bc +ac? shouldn' it be 8*sqrt(3)? Could you chek it again, plz??

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Could you show any algebraic, only algebraic solution, friend?

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Edited.

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The point "O" is known as "Fermat" or Torricelli Point of the triangle.

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what will be the algebric solution Help me

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The solution could be algebraical or geometrical, Thanks in advance for help!

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