1. Problem 9: Calculate the wingspan (length of segment DC) of the Wawa Goose given points A and B 4m apart, angles \(\angle BAC = 29^\circ\), \(\angle DAB = 81^\circ\), and \(\angle DBA = 47^\circ\). 2. First, find the length of segment AD using triangle ABD. We know \(AB = 4m\), \(\angle DAB = 81^\circ\), and \(\angle DBA = 47^\circ\). The third angle \(\angle ADB = 180^\circ - 81^\circ - 47^\circ = 52^\circ\). 3. Use the Law of Sines in triangle ABD: $$\frac{AD}{\sin 47^\circ} = \frac{4}{\sin 52^\circ}$$ Solve for \(AD\): $$AD = \frac{4 \sin 47^\circ}{\sin 52^\circ}$$ 4. Calculate \(AD\) numerically: $$AD \approx \frac{4 \times 0.7314}{0.7880} \approx 3.71\,m$$ 5. Next, find length AC using triangle ABC. We know \(AB = 4m\), \(\angle BAC = 29^\circ\), and \(\angle ABC = 180^\circ - \angle DBA = 180^\circ - 47^\circ = 133^\circ\) because points A, B, C form a triangle and \(\angle ABC\) complements \(\angle DBA\). 6. Find \(\angle ACB = 180^\circ - 29^\circ - 133^\circ = 18^\circ\). 7. Use Law of Sines in triangle ABC: $$\frac{AC}{\sin 133^\circ} = \frac{4}{\sin 18^\circ}$$ Solve for \(AC\): $$AC = \frac{4 \sin 133^\circ}{\sin 18^\circ}$$ 8. Calculate \(AC\) numerically: $$AC \approx \frac{4 \times 0.7431}{0.3090} \approx 9.62\,m$$ 9. Now, find the wingspan \(DC\). We know lengths \(AD \approx 3.71m\) and \(AC \approx 9.62m\), and the angle between them at A is \(\angle DAC = 180^\circ - (81^\circ + 29^\circ) = 70^\circ\). 10. Use Law of Cosines in triangle ADC: $$DC^2 = AD^2 + AC^2 - 2 \times AD \times AC \times \cos 70^\circ$$ 11. Calculate \(DC\): $$DC = \sqrt{3.71^2 + 9.62^2 - 2 \times 3.71 \times 9.62 \times \cos 70^\circ}$$ 12. Numerically evaluate: $$DC = \sqrt{13.76 + 92.54 - 2 \times 3.71 \times 9.62 \times 0.3420} = \sqrt{106.3 - 24.44} = \sqrt{81.86} \approx 9.05\,m$$ --- 13. Problem 10: Calculate the total distance the helicopter travels from Cameronville (C) to point B across the bay. 14. The helicopter flies from C at \(75^\circ\) east of north until it is due north of B, then turns directly to B. The point B is 10 km from Cameronville at \(20^\circ\) south of east. 15. Represent the problem as triangle with points C (start), H (point due north of B on helicopter path), and B. 16. The angle at C between the helicopter's initial path and the line CB is \(75^\circ + 20^\circ = 95^\circ\). 17. The distance CB is 10 km. 18. Use Law of Sines in triangle CHB: $$\frac{CH}{\sin 20^\circ} = \frac{10}{\sin 75^\circ}$$ Solve for \(CH\): $$CH = \frac{10 \sin 20^\circ}{\sin 75^\circ}$$ 19. Calculate \(CH\) numerically: $$CH \approx \frac{10 \times 0.3420}{0.9659} \approx 3.54\,km$$ 20. The helicopter flies distance CH first, then HB directly to B. 21. Find HB using Pythagoras since HB is south from H to B, and the helicopter turns directly to B. 22. The angle at H is \(90^\circ\) because H is due north of B. 23. Use Law of Cosines or right triangle properties: $$HB = 10 \times \sin 75^\circ \approx 10 \times 0.9659 = 9.66\,km$$ 24. Total distance traveled by helicopter: $$CH + HB = 3.54 + 9.66 = 13.20\,km$$ Final answers: - Wingspan of Wawa Goose \(DC \approx 9.05\,m\) - Total helicopter distance \(\approx 13.20\,km\)