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Question

In a ABC with usual notations, if tanA,tanB satisfy the equation 3x24x+3<0 and (ab)2>c2ab, then number of such triangles possible is

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Solution

Given, (ab)2>c2ab ...(1)
3x24x+3<03x23xx+3<0(3x1)(x3)<013<x<313<tanA,tanB<3π6<A,B<π3π3<C<2π312<cosC<1212<a2+b2c22ab<12ab<a2+b2c2<abc2ab<(ab)2+2ab<c2+abc23ab<(ab)2<c2ab ...(2)

Since, eqn (1) and (2) are contradict each others.
Therefore, no such triangle is possible.

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