A triangle has two of its sides measuring 6 cm and 8 cm, and the triangle is obtuse-angled. For how many whole-number values of the third side, in centimetres, is this possible?

Call the third side $x$. The triangle inequality needs 8 − 6 < $x$ < 8 + 6, so $x$ runs from 3 to 13. The obtuse angle always faces the longest side, so there are two cases. If $x$ is the longest, obtuseness needs $x^2>6^2+8^2=100$, so $x$ > 10, giving $x=11$, 12, 13. If 8 is the longest, it needs $8^2>6^2+x^2$, so $x^2<28$ and $x$ ≤ 5, giving $x=3$, 4, 5. Altogether 6 values. Note $x=10$ is excluded because 6-8-10 is right-angled, not obtuse, and $x=6$ to 9 give acute triangles. The trap 3 counts only one of the two branches, 11 counts every triangle and ignores the obtuse condition, and 10 is that count with only the single right-angled case removed.