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Number of obtuse angled triangles with integral sides whose longest side has length 13 units.

Is there any theorem or approach to solve such problems.

I tried to use the triangle theorem which states that if 2 sides of a triangle are known the third side can have values ranging from difference of the given sides to the addition of these sides.

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An approach:

WLOG let $a\geqslant b$, then we have $0< b \leqslant a < 13$, $a+b > 13$ and $a^2+b^2< 169$. You may then want to count with $a = 12, 11, ...7$

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Hint:

From cosine law, we have $$a^2+b^2-2ab\cos\theta=13^2$$ Since $\theta$ is obtuse $\implies -1<cos\theta<0$. $$a^2+b^2<13^2<(a+b)^2$$

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