Source: 1976 Euclid Part A Problem 9
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A circle has an inscribed triangle whose sides are $5\sqrt{3}$, $10\sqrt{3}$, and $15$. The measure of the angle subtended at the centre of the circle by the shortest side is 

$\textbf{(A) } 30 \qquad \textbf{(B) } 45 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } \text{none of these}$
1. **Identify the sides of the triangle and their relationship:**
   The sides of the triangle are \(5\sqrt{3}\), \(10\sqrt{3}\), and \(15\). We need to determine the type of triangle and the angles involved.

2. **Check if the triangle is a right triangle:**
   To check if the triangle is a right triangle, we use the Pythagorean theorem. For a triangle with sides \(a\), \(b\), and \(c\) (where \(c\) is the hypotenuse), the relationship \(a^2 + b^2 = c^2\) must hold.
   \[
   (5\sqrt{3})^2 + (10\sqrt{3})^2 = 15^2
   \]
   \[
   75 + 300 = 225
   \]
   \[
   375 \neq 225
   \]
   Since the Pythagorean theorem does not hold, the triangle is not a right triangle.

3. **Calculate the area using Heron's formula:**
   Heron's formula for the area of a triangle with sides \(a\), \(b\), and \(c\) is:
   \[
   s = \frac{a + b + c}{2}
   \]
   \[
   s = \frac{5\sqrt{3} + 10\sqrt{3} + 15}{2} = \frac{15\sqrt{3} + 15}{2} = \frac{15(\sqrt{3} + 1)}{2}
   \]
   The area \(A\) is:
   \[
   A = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   \[
   A = \sqrt{\frac{15(\sqrt{3} + 1)}{2} \left( \frac{15(\sqrt{3} + 1)}{2} - 5\sqrt{3} \right) \left( \frac{15(\sqrt{3} + 1)}{2} - 10\sqrt{3} \right) \left( \frac{15(\sqrt{3} + 1)}{2} - 15 \right)}
   \]
   Simplifying the terms inside the square root:
   \[
   A = \sqrt{\frac{15(\sqrt{3} + 1)}{2} \left( \frac{15(\sqrt{3} + 1) - 10\sqrt{3}}{2} \right) \left( \frac{15(\sqrt{3} + 1) - 20\sqrt{3}}{2} \right) \left( \frac{15(\sqrt{3} + 1) - 30}{2} \right)}
   \]
   \[
   A = \sqrt{\frac{15(\sqrt{3} + 1)}{2} \left( \frac{15 + 15\sqrt{3} - 10\sqrt{3}}{2} \right) \left( \frac{15 + 15\sqrt{3} - 20\sqrt{3}}{2} \right) \left( \frac{15 + 15\sqrt{3} - 30}{2} \right)}
   \]
   \[
   A = \sqrt{\frac{15(\sqrt{3} + 1)}{2} \left( \frac{15 + 5\sqrt{3}}{2} \right) \left( \frac{15 - 5\sqrt{3}}{2} \right) \left( \frac{15 - 15}{2} \right)}
   \]
   \[
   A = \sqrt{\frac{15(\sqrt{3} + 1)}{2} \cdot \frac{15 + 5\sqrt{3}}{2} \cdot \frac{15 - 5\sqrt{3}}{2} \cdot 0}
   \]
   Since one of the terms is zero, the area calculation is incorrect. We need to re-evaluate the area using a different approach.

4. **Re-evaluate the triangle type:**
   The sides \(5\sqrt{3}\), \(10\sqrt{3}\), and \(15\) suggest a 30-60-90 triangle. In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\).

5. **Determine the angles:**
   In a 30-60-90 triangle, the angles are 30°, 60°, and 90°. The shortest side \(5\sqrt{3}\) is opposite the 30° angle.

6. **Calculate the angle subtended at the center:**
   The angle subtended at the center of the circle by the shortest side is twice the angle at the vertex of the triangle.
   \[
   2 \times 30° = 60°
   \]

The final answer is \(\boxed{60}\).