Helicopter Bearings 190F46

1. **Problem Statement:** Two helicopters start from the same point. The first flies 5 km on a bearing of $60^\circ$, the second flies 7 km on a bearing of $130^\circ$. We need to find: a. The quadrant of the first helicopter's bearing. b. The quadrant of the second helicopter's bearing. c. The angle between their paths. d. Illustrate and label the paths. e. Calculate the distance between the two helicopters after flying. 2. **Understanding Bearings and Quadrants:** - Bearings are measured clockwise from the north. - Quadrants are divided as: - Quadrant 1: $0^\circ$ to $90^\circ$ - Quadrant 2: $90^\circ$ to $180^\circ$ - Quadrant 3: $180^\circ$ to $270^\circ$ - Quadrant 4: $270^\circ$ to $360^\circ$ 3. **Step a: Quadrant of first helicopter bearing $60^\circ$** - Since $60^\circ$ is between $0^\circ$ and $90^\circ$, the first helicopter is flying in **Quadrant 1**. 4. **Step b: Quadrant of second helicopter bearing $130^\circ$** - Since $130^\circ$ is between $90^\circ$ and $180^\circ$, the second helicopter is flying in **Quadrant 2**. 5. **Step c: Angle between the two paths** - The angle between the paths is the difference between the bearings: $$\theta = 130^\circ - 60^\circ = 70^\circ$$ 6. **Step d: Illustration and labeling** - The first helicopter path is a vector 5 km long at $60^\circ$ from north. - The second helicopter path is a vector 7 km long at $130^\circ$ from north. 7. **Step e: Distance between the two helicopters** - Use the Law of Cosines to find the distance $d$ between the helicopters: $$d^2 = 5^2 + 7^2 - 2 \times 5 \times 7 \times \cos(70^\circ)$$ $$d^2 = 25 + 49 - 70 \times \cos(70^\circ)$$ Calculate $\cos(70^\circ) \approx 0.3420$: $$d^2 = 74 - 70 \times 0.3420 = 74 - 23.94 = 50.06$$ $$d = \sqrt{50.06} \approx 7.07 \text{ km}$$ **Final answer:** - a. First helicopter is in Quadrant 1. - b. Second helicopter is in Quadrant 2. - c. Angle between paths is $70^\circ$. - e. Distance between helicopters is approximately $7.07$ km.