1. **Find q in triangle PQR** Given that triangle PQR is right-angled at Q, with side PQ = 4, angle R = 45°, and hypotenuse PR = q. 2. Since angle R is 45°, and the triangle is right-angled at Q, angle P must also be 45° (sum of angles in triangle is 180°). 3. Triangle PQR is therefore a 45°-45°-90° triangle, which means the legs are equal and the hypotenuse is $\sqrt{2}$ times a leg. 4. Since PQ = 4, the other leg QR is also 4. 5. The hypotenuse PR = $q = 4 \times \sqrt{2} = 4\sqrt{2}$. 6. **Calculate length of YZ in isosceles triangle XYZ** Given $|XY| = |XZ| = 6$ cm and angle $YXZ = 120^\circ$. 7. By the Law of Cosines: $$YZ^2 = XY^2 + XZ^2 - 2 \times XY \times XZ \times \cos(120^\circ)$$ $$YZ^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos(120^\circ)$$ $$YZ^2 = 36 + 36 - 72 \times (-0.5) = 72 + 36 = 108$$ 8. Therefore, $$YZ = \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \approx 10.39 \text{ cm}$$ 9. **Calculate the slant height of the cone** Given height $h = 42$ cm and semi-vertical angle $30^\circ$. 10. The slant height $l$ relates to height by $$\tan(30^\circ) = \frac{\text{radius}}{42}$$ 11. Since slant height $l = \frac{42}{\cos(30^\circ)}$ 12. Calculate: $$\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$$ $$l = \frac{42}{0.866} \approx 48.47 \text{ cm}$$ 13. **Find distance between two ships** From tower height $60$ m with angles of depression $45^\circ$ and $30^\circ$. 14. Distance to ship 1: $$d_1 = 60 \times \tan(45^\circ) = 60 \times 1 = 60 \text{ m}$$ 15. Distance to ship 2: $$d_2 = 60 \times \tan(30^\circ) = 60 \times \frac{1}{\sqrt{3}} = 60 \times 0.577 = 34.64 \text{ m}$$ 16. Distance between ships is: $$|d_1 - d_2| = 60 - 34.64 = 25.36 \text{ m}$$