Answer :
Let's start by analyzing the given problem and using trigonometry to solve it.
Given:
1. Ship [tex]\( P \)[/tex] is 3 km due east of the harbor.
2. Ship [tex]\( Q \)[/tex] is also 3 km from the harbor but on a bearing of 042° from the harbor.
We need to:
1. Find the distance between the two ships.
2. Find the bearing of ship [tex]\( Q \)[/tex] from ship [tex]\( P \)[/tex].
Step 1: Determine the coordinates of Ships [tex]\( P \)[/tex] and [tex]\( Q \)[/tex].
The coordinates can be determined using basic trigonometric relations.
1. Ship [tex]\( P \)[/tex] is 3 km due east of the harbor. Therefore, the coordinates of ship [tex]\( P \)[/tex] relative to the harbor are:
[tex]\[ P = (3, 0) \][/tex]
2. Ship [tex]\( Q \)[/tex] is 3 km from the harbor on a bearing of 042°. In a standard coordinate system:
[tex]\[ Q_x = 3 \cdot \sin(42^\circ) \][/tex]
[tex]\[ Q_y = 3 \cdot \cos(42^\circ) \][/tex]
By approximating [tex]\( \sin(42^\circ) \)[/tex] and [tex]\( \cos(42^\circ) \)[/tex]:
[tex]\[ \sin(42^\circ) \approx 0.6691 \][/tex]
[tex]\[ \cos(42^\circ) \approx 0.7431 \][/tex]
Thus,
[tex]\[ Q_x = 3 \cdot 0.6691 \approx 2.0073 \text{ km} \][/tex]
[tex]\[ Q_y = 3 \cdot 0.7431 \approx 2.2293 \text{ km} \][/tex]
So, the coordinates of ship [tex]\( Q \)[/tex] are:
[tex]\[ Q \approx (2.0073, 2.2293) \][/tex]
Step 2: Calculate the distance between ships [tex]\( P \)[/tex] and [tex]\( Q \)[/tex].
Using the distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates we have:
[tex]\[ d = \sqrt{(2.0073 - 3)^2 + (2.2293 - 0)^2} \][/tex]
[tex]\[ d = \sqrt{(-0.9927)^2 + 2.2293^2} \][/tex]
[tex]\[ d = \sqrt{0.9855 + 4.9727} \][/tex]
[tex]\[ d = \sqrt{5.9582} \][/tex]
[tex]\[ d \approx 2.442 \text{ km} \][/tex]
So, the distance between the two ships is approximately [tex]\( 2.442 \)[/tex] km.
Step 3: Calculate the bearing of ship [tex]\( Q \)[/tex] from ship [tex]\( P \)[/tex].
The bearing can be found using trigonometric relationships. Specifically, the bearing can be calculated using the inverse tangent function:
[tex]\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \][/tex]
Where:
[tex]\[ x = Q_x - P_x = 2.0073 - 3 = -0.9927 \][/tex]
[tex]\[ y = Q_y - P_y = 2.2293 - 0 = 2.2293 \][/tex]
Thus:
[tex]\[ \theta = \tan^{-1}\left(\frac{2.2293}{-0.9927}\right) \][/tex]
Calculating the angle:
[tex]\[ \theta = \tan^{-1}(-2.245) \approx -66.94^\circ \][/tex]
Bearings are usually measured clockwise from the north. Therefore, we must adjust this angle to fit within the 0° to 360° range:
[tex]\[ \theta_{corrected} = 360^\circ - 66.94^\circ = 293.06^\circ \][/tex]
Therefore, the bearing of ship [tex]\( Q \)[/tex] from ship [tex]\( P \)[/tex] is approximately [tex]\( 293.06^\circ \)[/tex].
Given:
1. Ship [tex]\( P \)[/tex] is 3 km due east of the harbor.
2. Ship [tex]\( Q \)[/tex] is also 3 km from the harbor but on a bearing of 042° from the harbor.
We need to:
1. Find the distance between the two ships.
2. Find the bearing of ship [tex]\( Q \)[/tex] from ship [tex]\( P \)[/tex].
Step 1: Determine the coordinates of Ships [tex]\( P \)[/tex] and [tex]\( Q \)[/tex].
The coordinates can be determined using basic trigonometric relations.
1. Ship [tex]\( P \)[/tex] is 3 km due east of the harbor. Therefore, the coordinates of ship [tex]\( P \)[/tex] relative to the harbor are:
[tex]\[ P = (3, 0) \][/tex]
2. Ship [tex]\( Q \)[/tex] is 3 km from the harbor on a bearing of 042°. In a standard coordinate system:
[tex]\[ Q_x = 3 \cdot \sin(42^\circ) \][/tex]
[tex]\[ Q_y = 3 \cdot \cos(42^\circ) \][/tex]
By approximating [tex]\( \sin(42^\circ) \)[/tex] and [tex]\( \cos(42^\circ) \)[/tex]:
[tex]\[ \sin(42^\circ) \approx 0.6691 \][/tex]
[tex]\[ \cos(42^\circ) \approx 0.7431 \][/tex]
Thus,
[tex]\[ Q_x = 3 \cdot 0.6691 \approx 2.0073 \text{ km} \][/tex]
[tex]\[ Q_y = 3 \cdot 0.7431 \approx 2.2293 \text{ km} \][/tex]
So, the coordinates of ship [tex]\( Q \)[/tex] are:
[tex]\[ Q \approx (2.0073, 2.2293) \][/tex]
Step 2: Calculate the distance between ships [tex]\( P \)[/tex] and [tex]\( Q \)[/tex].
Using the distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates we have:
[tex]\[ d = \sqrt{(2.0073 - 3)^2 + (2.2293 - 0)^2} \][/tex]
[tex]\[ d = \sqrt{(-0.9927)^2 + 2.2293^2} \][/tex]
[tex]\[ d = \sqrt{0.9855 + 4.9727} \][/tex]
[tex]\[ d = \sqrt{5.9582} \][/tex]
[tex]\[ d \approx 2.442 \text{ km} \][/tex]
So, the distance between the two ships is approximately [tex]\( 2.442 \)[/tex] km.
Step 3: Calculate the bearing of ship [tex]\( Q \)[/tex] from ship [tex]\( P \)[/tex].
The bearing can be found using trigonometric relationships. Specifically, the bearing can be calculated using the inverse tangent function:
[tex]\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \][/tex]
Where:
[tex]\[ x = Q_x - P_x = 2.0073 - 3 = -0.9927 \][/tex]
[tex]\[ y = Q_y - P_y = 2.2293 - 0 = 2.2293 \][/tex]
Thus:
[tex]\[ \theta = \tan^{-1}\left(\frac{2.2293}{-0.9927}\right) \][/tex]
Calculating the angle:
[tex]\[ \theta = \tan^{-1}(-2.245) \approx -66.94^\circ \][/tex]
Bearings are usually measured clockwise from the north. Therefore, we must adjust this angle to fit within the 0° to 360° range:
[tex]\[ \theta_{corrected} = 360^\circ - 66.94^\circ = 293.06^\circ \][/tex]
Therefore, the bearing of ship [tex]\( Q \)[/tex] from ship [tex]\( P \)[/tex] is approximately [tex]\( 293.06^\circ \)[/tex].
