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Sagot :
Let's break down the problem step-by-step to find the distance between the two aircraft and the bearing of P from Q after three hours:
### Step 1: Calculate the distances traveled by each aircraft
Aircraft P:
- Speed: 26 km/h
- Time: 3 hours
[tex]\[ \text{Distance}_{\text{P}} = \text{Speed} \times \text{Time} = 26 \, \text{km/h} \times 3 \, \text{hours} = 78 \, \text{km} \][/tex]
Aircraft Q:
- Speed: 41 km/h
- Time: 3 hours
[tex]\[ \text{Distance}_{\text{Q}} = \text{Speed} \times \text{Time} = 41 \, \text{km/h} \times 3 \, \text{hours} = 123 \, \text{km} \][/tex]
### Step 2: Calculate the coordinates of each aircraft
To find the position of each aircraft after three hours, we need to calculate their coordinates (considering bearings are angles measured clockwise from North).
Aircraft P:
- Distance: 78 km
- Bearing: 345° (Clockwise from North)
Using trigonometric functions:
[tex]\[ x_{\text{P}} = -20.19 \, \text{km} \][/tex]
[tex]\[ y_{\text{P}} = 75.34 \, \text{km} \][/tex]
Aircraft Q:
- Distance: 123 km
- Bearing: 75° (Clockwise from North)
Using trigonometric functions:
[tex]\[ x_{\text{Q}} = 118.81 \, \text{km} \][/tex]
[tex]\[ y_{\text{Q}} = 31.83 \, \text{km} \][/tex]
### Step 3: Calculate the distance between the two aircraft
Using the Euclidean distance formula between points [tex]\((x_p, y_p)\)[/tex] and [tex]\((x_q, y_q)\)[/tex]:
[tex]\[ \text{Distance}_{\text{between}} = \sqrt{(x_{\text{Q}} - x_{\text{P}})^2 + (y_{\text{Q}} - y_{\text{P}})^2} \][/tex]
Plugging in the coordinates:
[tex]\[ \text{Distance}_{\text{between}} = \sqrt{(118.81 - (-20.19))^2 + (31.83 - 75.34)^2} = 145.65 \, \text{km} \][/tex]
### Step 4: Calculate the bearing from Q to P
To find the bearing from Q to P:
- Calculate the change in coordinates:
[tex]\[ \Delta x = x_{\text{P}} - x_{\text{Q}} = -20.19 - 118.81 = -138.99 \][/tex]
[tex]\[ \Delta y = y_{\text{P}} - y_{\text{Q}} = 75.34 - 31.83 = 43.51 \][/tex]
- Calculate the bearing using the arctangent of the differences:
[tex]\[ \text{Bearing} = \tan^{-1}\left(\frac{\Delta x}{\Delta y}\right) \][/tex]
The arctangent in degrees:
[tex]\[ \text{Bearing}_{\text{deg}} = 287.38^\circ \][/tex]
### Summary
(a) The distance between the two aircraft after three hours is approximately 146 km (to the nearest kilometer).
(b) The bearing of P from Q after three hours is approximately 287° (to the nearest degree).
### Step 1: Calculate the distances traveled by each aircraft
Aircraft P:
- Speed: 26 km/h
- Time: 3 hours
[tex]\[ \text{Distance}_{\text{P}} = \text{Speed} \times \text{Time} = 26 \, \text{km/h} \times 3 \, \text{hours} = 78 \, \text{km} \][/tex]
Aircraft Q:
- Speed: 41 km/h
- Time: 3 hours
[tex]\[ \text{Distance}_{\text{Q}} = \text{Speed} \times \text{Time} = 41 \, \text{km/h} \times 3 \, \text{hours} = 123 \, \text{km} \][/tex]
### Step 2: Calculate the coordinates of each aircraft
To find the position of each aircraft after three hours, we need to calculate their coordinates (considering bearings are angles measured clockwise from North).
Aircraft P:
- Distance: 78 km
- Bearing: 345° (Clockwise from North)
Using trigonometric functions:
[tex]\[ x_{\text{P}} = -20.19 \, \text{km} \][/tex]
[tex]\[ y_{\text{P}} = 75.34 \, \text{km} \][/tex]
Aircraft Q:
- Distance: 123 km
- Bearing: 75° (Clockwise from North)
Using trigonometric functions:
[tex]\[ x_{\text{Q}} = 118.81 \, \text{km} \][/tex]
[tex]\[ y_{\text{Q}} = 31.83 \, \text{km} \][/tex]
### Step 3: Calculate the distance between the two aircraft
Using the Euclidean distance formula between points [tex]\((x_p, y_p)\)[/tex] and [tex]\((x_q, y_q)\)[/tex]:
[tex]\[ \text{Distance}_{\text{between}} = \sqrt{(x_{\text{Q}} - x_{\text{P}})^2 + (y_{\text{Q}} - y_{\text{P}})^2} \][/tex]
Plugging in the coordinates:
[tex]\[ \text{Distance}_{\text{between}} = \sqrt{(118.81 - (-20.19))^2 + (31.83 - 75.34)^2} = 145.65 \, \text{km} \][/tex]
### Step 4: Calculate the bearing from Q to P
To find the bearing from Q to P:
- Calculate the change in coordinates:
[tex]\[ \Delta x = x_{\text{P}} - x_{\text{Q}} = -20.19 - 118.81 = -138.99 \][/tex]
[tex]\[ \Delta y = y_{\text{P}} - y_{\text{Q}} = 75.34 - 31.83 = 43.51 \][/tex]
- Calculate the bearing using the arctangent of the differences:
[tex]\[ \text{Bearing} = \tan^{-1}\left(\frac{\Delta x}{\Delta y}\right) \][/tex]
The arctangent in degrees:
[tex]\[ \text{Bearing}_{\text{deg}} = 287.38^\circ \][/tex]
### Summary
(a) The distance between the two aircraft after three hours is approximately 146 km (to the nearest kilometer).
(b) The bearing of P from Q after three hours is approximately 287° (to the nearest degree).
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