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To solve the problem of finding the bearing of point [tex]\( Q \)[/tex] from point [tex]\( P \)[/tex] after the man has walked two legs of his journey, we can break down the solution into several steps that involve converting bearings to vector components, adding those vectors, and reconverting the result back into a bearing. Here is a detailed, step-by-step solution:
### Step 1: Represent each leg of the journey as vectors
Firstly, we need to convert the bearings and distances of each leg of the journey into Cartesian coordinates (x, y).
#### Leg 1: Walking 4 km at a bearing of [tex]\( 023^{\circ} \)[/tex]
1. Distance [tex]\( r_1 = 4 \)[/tex] km
2. Bearing [tex]\( \theta_1 = 23^{\circ} \)[/tex]
3. Convert the angle [tex]\( \theta_1 \)[/tex] to radians [tex]\( \theta_1 = \frac{23 \pi}{180} \)[/tex]
4. Calculate the x and y components:
- [tex]\( x_1 = r_1 \cos(\theta_1) = 4 \cos(\frac{23 \pi}{180}) = 3.682 \)[/tex]
- [tex]\( y_1 = r_1 \sin(\theta_1) = 4 \sin(\frac{23 \pi}{180}) = 1.563 \)[/tex]
#### Leg 2: Walking 4 km at a bearing of [tex]\( 113^{\circ} \)[/tex]
1. Distance [tex]\( r_2 = 4 \)[/tex] km
2. Bearing [tex]\( \theta_2 = 113^{\circ} \)[/tex]
3. Convert the angle [tex]\( \theta_2 \)[/tex] to radians [tex]\( \theta_2 = \frac{113 \pi}{180} \)[/tex]
4. Calculate the x and y components:
- [tex]\( x_2 = r_2 \cos(\theta_2) = 4 \cos(\frac{113 \pi}{180}) = -1.563 \)[/tex]
- [tex]\( y_2 = r_2 \sin(\theta_2) = 4 \sin(\frac{113 \pi}{180}) = 3.682 \)[/tex]
### Step 2: Add the vectors to find the resulting position
Combine the x and y components of the two legs to get the total displacement:
1. [tex]\( x_\text{total} = x_1 + x_2 = 3.682 + (-1.563) = 2.119 \)[/tex]
2. [tex]\( y_\text{total} = y_1 + y_2 = 1.563 + 3.682 = 5.245 \)[/tex]
### Step 3: Convert the resulting position back to polar coordinates
To find the resulting distance [tex]\( r \)[/tex] and bearing [tex]\( \theta \)[/tex]:
1. Calculate the distance [tex]\( r \)[/tex]:
- [tex]\( r = \sqrt{x_\text{total}^2 + y_\text{total}^2} = \sqrt{2.119^2 + 5.245^2} \approx 5.657 \)[/tex] km
2. Calculate the bearing [tex]\( \theta \)[/tex]:
- [tex]\( \theta = \arctan\left(\frac{y_\text{total}}{x_\text{total}}\right) = \arctan\left(\frac{5.245}{2.119}\right) \approx 68^{\circ} \)[/tex]
### Step 4: Normalize the bearing
Since the bearing [tex]\( \theta \)[/tex] is positive and less than 360 degrees, it remains as it is. Therefore, the bearing of point [tex]\( Q \)[/tex] from point [tex]\( P \)[/tex] is:
[tex]\[ \boxed{68^{\circ}} \][/tex]
Thus, after walking 4 km on a bearing of [tex]\( 023^{\circ} \)[/tex] and then 4 km on a bearing of [tex]\( 113^{\circ} \)[/tex], the bearing of point [tex]\( Q \)[/tex] from point [tex]\( P \)[/tex] is approximately [tex]\( 68^{\circ} \)[/tex].
### Step 1: Represent each leg of the journey as vectors
Firstly, we need to convert the bearings and distances of each leg of the journey into Cartesian coordinates (x, y).
#### Leg 1: Walking 4 km at a bearing of [tex]\( 023^{\circ} \)[/tex]
1. Distance [tex]\( r_1 = 4 \)[/tex] km
2. Bearing [tex]\( \theta_1 = 23^{\circ} \)[/tex]
3. Convert the angle [tex]\( \theta_1 \)[/tex] to radians [tex]\( \theta_1 = \frac{23 \pi}{180} \)[/tex]
4. Calculate the x and y components:
- [tex]\( x_1 = r_1 \cos(\theta_1) = 4 \cos(\frac{23 \pi}{180}) = 3.682 \)[/tex]
- [tex]\( y_1 = r_1 \sin(\theta_1) = 4 \sin(\frac{23 \pi}{180}) = 1.563 \)[/tex]
#### Leg 2: Walking 4 km at a bearing of [tex]\( 113^{\circ} \)[/tex]
1. Distance [tex]\( r_2 = 4 \)[/tex] km
2. Bearing [tex]\( \theta_2 = 113^{\circ} \)[/tex]
3. Convert the angle [tex]\( \theta_2 \)[/tex] to radians [tex]\( \theta_2 = \frac{113 \pi}{180} \)[/tex]
4. Calculate the x and y components:
- [tex]\( x_2 = r_2 \cos(\theta_2) = 4 \cos(\frac{113 \pi}{180}) = -1.563 \)[/tex]
- [tex]\( y_2 = r_2 \sin(\theta_2) = 4 \sin(\frac{113 \pi}{180}) = 3.682 \)[/tex]
### Step 2: Add the vectors to find the resulting position
Combine the x and y components of the two legs to get the total displacement:
1. [tex]\( x_\text{total} = x_1 + x_2 = 3.682 + (-1.563) = 2.119 \)[/tex]
2. [tex]\( y_\text{total} = y_1 + y_2 = 1.563 + 3.682 = 5.245 \)[/tex]
### Step 3: Convert the resulting position back to polar coordinates
To find the resulting distance [tex]\( r \)[/tex] and bearing [tex]\( \theta \)[/tex]:
1. Calculate the distance [tex]\( r \)[/tex]:
- [tex]\( r = \sqrt{x_\text{total}^2 + y_\text{total}^2} = \sqrt{2.119^2 + 5.245^2} \approx 5.657 \)[/tex] km
2. Calculate the bearing [tex]\( \theta \)[/tex]:
- [tex]\( \theta = \arctan\left(\frac{y_\text{total}}{x_\text{total}}\right) = \arctan\left(\frac{5.245}{2.119}\right) \approx 68^{\circ} \)[/tex]
### Step 4: Normalize the bearing
Since the bearing [tex]\( \theta \)[/tex] is positive and less than 360 degrees, it remains as it is. Therefore, the bearing of point [tex]\( Q \)[/tex] from point [tex]\( P \)[/tex] is:
[tex]\[ \boxed{68^{\circ}} \][/tex]
Thus, after walking 4 km on a bearing of [tex]\( 023^{\circ} \)[/tex] and then 4 km on a bearing of [tex]\( 113^{\circ} \)[/tex], the bearing of point [tex]\( Q \)[/tex] from point [tex]\( P \)[/tex] is approximately [tex]\( 68^{\circ} \)[/tex].
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