Let's break down the problem step by step: ### Given: - **Airplane's airspeed** = 230 mph (relative to the air) - **Airplane's bearing** = 285° - **Wind speed** = 35 mph - **Wind bearing** = 260° We are to find: - The **plane's true bearing** (direction of travel relative to north) - The **groundspeed** (speed relative to the ground) --- ## 1. **Convert Bearings to Standard Angles** Bearings are measured clockwise from north (0° = north, 90° = east, 180° = south, 270° = west). ### a) Airplane's velocity (relative to air): - Bearing = 285° - From north, clockwise, so - \( \theta_{plane} = 285° \) ### b) Wind velocity: - Bearing = 260° - \( \theta_{wind} = 260° \) --- ## 2. **Find Components** Let’s express both vectors in **rectangular coordinates** (East = x, North = y): For bearing \( \theta \), the components are: - \( x = V \sin(\theta) \) - \( y = V \cos(\theta) \) #### a) Airplane (relative to air): - \( V_{plane} = 230 \) mph, \( \theta_{plane} = 285° \) \[ \sin 285° = \sin(-75°) = -0.9659 \] \[ \cos 285° = \cos(-75°) = 0.2588 \] So, \[ V_{plane,x} = 230 \times \sin(285°) = 230 \times (-0.9659) = -222.157 \text{ mph} \] \[ V_{plane,y} = 230 \times \cos(285°) = 230 \times 0.2588 = 59.524 \text{ mph} \] #### b) Wind: - \( V_{wind} = 35 \) mph, \( \theta_{wind} = 260° \) \[ \sin 260° = \sin(-100°) = -0.9848 \] \[ \cos 260° = \cos(-100°) = -0.1736 \] So, \[ V_{wind,x} = 35 \times \sin(260°) = 35 \times (-0.9848) = -34.468 \text{ mph} \] \[ V_{wind,y} = 35 \times \cos(260°) = 35 \times (-0.1736) = -6.076 \text{ mph} \] --- ## 3. **Ground Velocity** Add components: \[ V_{ground,x} = V_{plane,x} + V_{wind,x} = -222.157 + (-34.468) = -256.625 \text{ mph} \] \[ V_{ground,y} = V_{plane,y} + V_{wind,y} = 59.524 + (-6.076) = 53.448 \text{ mph} \] --- ## 4. **Groundspeed** \[ \text{Groundspeed} = \sqrt{(-256.625)^2 + (53.448)^2} \] \[ = \sqrt{65860.3 + 2858.7} \] \[ = \sqrt{68719} \] \[ \approx 262.3 \text{ mph} \] --- ## 5. **True Bearing** The true bearing is measured **clockwise from north**. First, find the angle θ from the positive y-axis (north): \[ \tan \theta = \frac{|x|}{|y|} = \frac{256.625}{53.448} = 4.8 \] \[ \theta = \arctan(4.8) \approx 78.3^\circ \] But since x is negative and y is positive, the vector is in the **northwest** quadrant (second quadrant). Bearings are measured clockwise from north. - The angle west of north is 78.3° - So the bearing is \( 360° - 78.3° = 281.7° \) **Alternatively**, since x is negative and y is positive, the vector is west of north. So, the **true bearing** is: \[ \boxed{281.7^\circ} \] --- ## **Final Answers** ### **Groundspeed:** \[ \boxed{262.3 \text{ mph}} \] ### **True Bearing:** \[ \boxed{282^\circ} \] (Rounded to the nearest degree.) --- ### **Summary Table** | Quantity | Value | |---------------|----------------| | Groundspeed | 262.3 mph | | True Bearing | 282° |