1. **Problem statement:** Determine the length of chord $PQ$ in the circle with center $C$, where $CP = CQ = 18$ m and the angle $\angle RPQ = 15^\circ$. The segment $RP = 35$ m. 2. **Understanding the problem:** We have an isosceles triangle $CPQ$ with $CP = CQ = 18$ m. The chord $PQ$ is the base of this triangle. We want to find the length of $PQ$. 3. **Formula used:** In an isosceles triangle with two equal sides $r$ and the angle between them $\theta$, the length of the base (chord) $PQ$ is given by: $$PQ = 2r \sin\left(\frac{\theta}{2}\right)$$ 4. **Identify the angle at center $C$ subtended by chord $PQ$:** Since $\angle RPQ = 15^\circ$ is given, but we need the central angle $\angle PCQ$ subtended by chord $PQ$. 5. **Using triangle $RPQ$ to find $\angle PCQ$:** Given $RP = 35$ m and $\angle RPQ = 15^\circ$, but this information is not directly related to the central angle $\angle PCQ$. 6. **Assuming $\angle PCQ = 2 \times 15^\circ = 30^\circ$ (common in circle geometry where the angle at the center is twice the angle at the circumference subtending the same chord):** 7. **Calculate chord length $PQ$:** $$PQ = 2 \times 18 \times \sin\left(\frac{30^\circ}{2}\right) = 36 \times \sin(15^\circ)$$ 8. **Calculate $\sin(15^\circ)$:** $$\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} \approx 0.2588$$ 9. **Final calculation:** $$PQ = 36 \times 0.2588 = 9.3168$$ 10. **Answer rounded to nearest tenth:** $$PQ \approx 9.3 \text{ metres}$$