Let's break down the positions of the villages step by step and then find the required distance and bearing from D to A.
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Position of A: We can consider village A to be at the origin, so we have:
- \( A(0, 0) \)
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Position of B: Since B is 4 km due east of A, the coordinates of B are:
- \( B(4, 0) \)
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Position of C: C is 3 km due south of B, giving us the coordinates:
- \( B(4, 0) \rightarrow C(4, -3) \)
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Position of D: D is 4 km in the direction S50°W from C. To calculate the coordinates of D, we need to determine the component form of the vector.
The direction S50°W means that we are heading 50 degrees towards the west from the south. This can be visualized as a standard position angle of \( 180° + 50° = 230° \) in standard position.
We can find the x and y components of the position vector for D using: \[ D_x = C_x + r \cdot \cos(230°) \] \[ D_y = C_y + r \cdot \sin(230°) \] where \( r = 4 \) km.
Calculating:
- \( C_x = 4 \)
- \( C_y = -3 \)
- \( \cos(230°) = -\cos(50°) \) and \( \sin(230°) = -\sin(50°) \)
- \( \cos(50°) \approx 0.6428 \) and \( \sin(50°) \approx 0.7660 \)
Thus: \[ D_x = 4 + 4 \cdot (-0.6428) = 4 - 2.5712 \approx 1.4288 \] \[ D_y = -3 + 4 \cdot (-0.7660) = -3 - 3.064 \approx -6.064 \] So the coordinates of D are approximately:
- \( D(1.4288, -6.064) \)
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Calculating distance from D to A: We can calculate the distance AD using the distance formula: \[ AD = \sqrt{(D_x - A_x)^2 + (D_y - A_y)^2} \] \[ AD = \sqrt{(1.4288 - 0)^2 + (-6.064 - 0)^2} \] \[ = \sqrt{(1.4288)^2 + (-6.064)^2} \] \[ = \sqrt{2.0428 + 36.8077} \approx \sqrt{38.8505} \approx 6.23 \text{ km} \]
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Calculating bearing from D to A: The bearing can be calculated using the tangent of the angle based on the coordinates: \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-6.064)}{0 - 1.4288} = \frac{6.064}{-1.4288} \] The angle θ from the positive x-axis (towards the east) is: \[ \theta = \tan^{-1}\left( \frac{6.064}{-1.4288} \right) \] Calculating: \[ \theta \approx -1.241 \text{ radians} \approx -71.1° + 180° (since we are in the second quadrant) \] \[ \theta \approx 108.9° \] Thus, the bearing is: \[ 108.9° \text{ or } 108.9° \text{ (clockwise from North)} \] Adjusting for that: \[ \text{Bearing from D to A} = 108.9° - 90° = 18.9° (\text{angle with North}) \]
Thus, the final results are:
- Distance from D to A: approximately 6.23 km
- Bearing from D to A: approximately S18.9°E or 108.9° from North.