1. **Problem:** Determine which sets of numbers can be the lengths of the sides of an obtuse-angled triangle. 2. **Formula and rule:** For a triangle with sides $a$, $b$, and $c$ where $c$ is the longest side, the triangle is obtuse-angled if: $$c^2 > a^2 + b^2$$ 3. **Check each set:** (a) $27, 10, 24$ (longest side $27$): $$27^2 = 729$$ $$10^2 + 24^2 = 100 + 576 = 676$$ Since $729 > 676$, this set can form an obtuse triangle. (b) $1.5, 2, 2.5$ (longest side $2.5$): $$2.5^2 = 6.25$$ $$1.5^2 + 2^2 = 2.25 + 4 = 6.25$$ Since $6.25 = 6.25$, this is a right triangle, not obtuse. (c) $7, 7, 7$ (equilateral): All sides equal, so all angles are $60^ ext{o}$, no obtuse angle. (d) $10\sqrt{3}, 10, 20$ (longest side $20$): $$20^2 = 400$$ $$ (10\sqrt{3})^2 + 10^2 = 100 \times 3 + 100 = 300 + 100 = 400$$ Since $400 = 400$, this is a right triangle, not obtuse. **Final answer:** Only set (a) can be the lengths of an obtuse-angled triangle.