Let's break down the problem step by step and **draw a diagram** to visualize the situation. --- ### **Step 1: Draw the Diagram** #### **1. First Leg: 90 km, 25° south of west** - Start at the origin (headquarters). - Draw a line 90 km at **25° south of west**. #### **2. Second Leg: 135 km, 70° west of north** - From the end of the first leg, draw a line 135 km at **70° west of north**. #### **3. Resultant Vector** - The direct path back to headquarters is the resultant vector from the endpoint back to the origin. --- ### **Step 2: Break Each Leg into Components** Let’s use the standard coordinate system: - **East (x) is positive** - **North (y) is positive** #### **Leg 1: 90 km at 25° south of west** - "West" is negative x-direction. - "25° south of west" means 25° below the negative x-axis. **Components:** - \( x_1 = -90 \cos(25^\circ) \) - \( y_1 = -90 \sin(25^\circ) \) #### **Leg 2: 135 km at 70° west of north** - "North" is positive y-direction. - "70° west of north" means 70° to the left of the y-axis. **Components:** - \( x_2 = -135 \sin(70^\circ) \) - \( y_2 = 135 \cos(70^\circ) \) --- ### **Step 3: Add the Components** **Total x-component:** \( x_{total} = x_1 + x_2 \) **Total y-component:** \( y_{total} = y_1 + y_2 \) --- ### **Step 4: Calculate the Distance Back** - The helicopter needs to fly from this point back to the origin. - The distance is the magnitude of the resultant vector: \[ d = \sqrt{(x_{total})^2 + (y_{total})^2} \] --- ### **Step 5: Find the Direction** - The direction is the angle **from the positive x-axis** (East), measured southwards (since both x and y will be negative). - Use: \[ \theta = \tan^{-1} \left(\frac{|y_{total}|}{|x_{total}|}\right) \] --- ## **Step-by-Step Calculation** ### **1. Calculate Components** #### **Leg 1:** - \( x_1 = -90 \cos(25^\circ) = -90 \times 0.9063 = -81.57 \) - \( y_1 = -90 \sin(25^\circ) = -90 \times 0.4226 = -38.03 \) #### **Leg 2:** - \( x_2 = -135 \sin(70^\circ) = -135 \times 0.9397 = -126.86 \) - \( y_2 = 135 \cos(70^\circ) = 135 \times 0.3420 = 46.17 \) --- ### **2. Add Components** - \( x_{total} = -81.57 + (-126.86) = -208.43 \) - \( y_{total} = -38.03 + 46.17 = 8.14 \) --- ### **3. Resultant Vector Back to Headquarters** The helicopter is at \((-208.43, 8.14)\). To return to headquarters, the vector is: - \( x = 208.43 \) (East) - \( y = -8.14 \) (South) --- ### **4. Distance** \[ d = \sqrt{(208.43)^2 + (8.14)^2} = \sqrt{43436.0 + 66.3} = \sqrt{43502.3} \approx 208.6 \text{ km} \] --- ### **5. Direction** \[ \theta = \tan^{-1} \left( \frac{8.14}{208.43} \right ) \approx \tan^{-1}(0.039) \approx 2.2^\circ \] This angle is **South of East** (since y-component is slightly south). --- ## **Final Answers** - **Distance:** **208.6 km** - **Direction:** **2.2° South of East** --- ### **Diagram Sketch:** ``` ^ | / | / | / | / | / (Second Leg: 135 km, 70° W of N) | / | / | / | / ---------|/--------------------------> / / / / / / (First Leg: 90 km, 25° S of W) ``` --- If you want a hand-drawn diagram, let me know!