The question involves two tangents, PQ and QR, drawn from an external point Q to a circle with center O. We are given the following data:

• OQ (distance from the center to the external point Q) = 41 cm
• ∠PQR = 60°
• Radius of the circle = 9 cm

We need to calculate the perimeter of ΔPRQ.

Let's solve this step-by-step:

1. From the property of tangents drawn from an external point to a circle, the lengths of tangents from a single external point are equal. Thus, PQ = QR.
2. In the triangle ΔOQP and ΔOQR, both are right triangles because the radius is perpendicular to the tangent at the point of contact. Hence, ∠OQP = ∠OQR = 90°.
3. We have OQ = 41 cm and OP = OR = 9 cm (radius).
4. Using the Pythagorean theorem in ΔOQP:
5. OP^2 + PQ^2 = OQ^2
6. Substituting the given values: 9^2 + PQ^2 = 41^2
7. Calculating further: 81 + PQ^2 = 1681
8. Therefore, PQ^2 = 1681 - 81 = 1600
9. It follows that PQ = \sqrt{1600} = 40 \text{ cm}
10. We have found that PQ = QR = 40 cm.
11. Now we calculate the perimeter of ΔPRQ:
12. Perimeter = PQ + QR + PR = 40 cm + 40 cm + PR
13. Considering ΔPQR where ∠PQR = 60°, we know PR is opposite this angle.
14. Since PQ = QR, ΔPQ is an isosceles triangle. Let PR = x:
15. We apply the cosine rule in ΔPQR: PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(60°)
16. Substitute the values: PR^2 = 40^2 + 40^2 - 2 \times 40 \times 40 \times \frac{1}{2}
17. Simplifying: PR^2 = 1600 + 1600 - 1600 = 1600
18. This gives us PR = \sqrt{1600} = 40 \text{ cm}
19. Finally, the Perimeter of ΔPRQ: 40 + 40 + 40 = 120 \text{ cm}

Therefore, the perimeter of ΔPRQ is 120 cm.

Hence, the correct answer is 120 cm.