1. **Problem statement:** An aircraft flies from A to B, distance AB = 600 km. Due to bad weather, it detours via C. The line CA makes an angle of 28° 15' with AB, and CA = 500 km. We need to find: (a) Distance CB (b) Angle ACB using the sine rule. 2. **Convert angle to decimal degrees:** 28° 15' = 28 + 15/60 = 28.25° 3. **Visualize the triangle:** Points A, B, and C form a triangle with sides: - AB = 600 km - CA = 500 km - CB = ? (to find) Angle between CA and AB is 28.25°. 4. **Use the Law of Cosines to find CB:** $$CB^2 = CA^2 + AB^2 - 2 \times CA \times AB \times \cos(28.25^\circ)$$ Calculate: $$CB^2 = 500^2 + 600^2 - 2 \times 500 \times 600 \times \cos(28.25^\circ)$$ $$CB^2 = 250000 + 360000 - 600000 \times \cos(28.25^\circ)$$ Calculate \(\cos(28.25^\circ)\): approximately 0.8819 $$CB^2 = 610000 - 600000 \times 0.8819 = 610000 - 529140 = 80860$$ $$CB = \sqrt{80860} \approx 284.3 \text{ km}$$ 5. **Use the Law of Sines to find angle ACB:** Label angles opposite sides: - Angle at A opposite CB - Angle at B opposite CA - Angle at C opposite AB We know: - Side AB = 600 km opposite angle C - Side CA = 500 km opposite angle B - Side CB = 284.3 km opposite angle A Using Law of Sines: $$\frac{\sin(ACB)}{AB} = \frac{\sin(28.25^\circ)}{CB}$$ Solve for \(\sin(ACB)\): $$\sin(ACB) = \frac{AB}{CB} \times \sin(28.25^\circ) = \frac{600}{284.3} \times \sin(28.25^\circ)$$ Calculate \(\sin(28.25^\circ)\): approximately 0.4735 $$\sin(ACB) = 2.11 \times 0.4735 = 1.0$$ Since sine cannot be more than 1, this means angle ACB is approximately 90°. 6. **Final answers:** (a) Distance CB is approximately 284.3 km. (b) Angle ACB is approximately 90°. This means the triangle is nearly right-angled at C.