1. The problem is to verify if the given triangle with sides $x = 8\sqrt{3}$, $y = 16\sqrt{3}$, and the side opposite the 30° angle being 8, forms a right-angled triangle with the right angle opposite side $x$. 2. Recall the Pythagorean theorem for right-angled triangles: $$a^2 + b^2 = c^2$$ where $c$ is the hypotenuse. 3. Here, the hypotenuse is $y = 16\sqrt{3}$, one leg is $x = 8\sqrt{3}$, and the other leg is 8. 4. Calculate the squares: $$x^2 = (8\sqrt{3})^2 = 8^2 \times (\sqrt{3})^2 = 64 \times 3 = 192$$ $$8^2 = 64$$ $$y^2 = (16\sqrt{3})^2 = 16^2 \times (\sqrt{3})^2 = 256 \times 3 = 768$$ 5. Check if $x^2 + 8^2 = y^2$: $$192 + 64 = 256$$ 6. Since $256 \neq 768$, the Pythagorean theorem does not hold with these values, so the triangle is not right-angled with $y$ as the hypotenuse and $x$ as a leg. 7. However, if the hypotenuse is $y = 16$, and the side opposite 30° is 8, then by the property of 30°-60°-90° triangles, the side opposite 30° is half the hypotenuse, which fits $8 = \frac{16}{2}$. 8. The side opposite 60° would be $8\sqrt{3}$, which matches $x = 8\sqrt{3}$. 9. Therefore, if $y = 16$ (not $16\sqrt{3}$), the triangle is a right-angled triangle with the right angle opposite side $x$. Final answer: The given values are not correct for a right-angled triangle as stated. The hypotenuse should be $16$, not $16\sqrt{3}$, for the triangle to be right-angled with the given sides and angles.