Answer :
Sure, let's break down the problem step-by-step to find the required distances.
### Setting Up the Problem
Given:
- The aeroplane flies from town [tex]\( X \)[/tex] to town [tex]\( Y \)[/tex] on a bearing of [tex]\( N 45^\circ E \)[/tex].
- Distance from [tex]\( X \)[/tex] to [tex]\( Y \)[/tex] is 200 km.
- Town [tex]\( Z \)[/tex] is directly east of town [tex]\( X \)[/tex].
- Y is on a line with bearing [tex]\( N 45^\circ E \)[/tex] from X, which means a 45-degree angle from the north towards the east.
First, we'll determine the coordinates of [tex]\( Y \)[/tex] relative to [tex]\( X \)[/tex]. Let's place [tex]\( X \)[/tex] at the origin [tex]\((0, 0)\)[/tex].
### Coordinates of [tex]\( Y \)[/tex]
To find the coordinates of [tex]\( Y \)[/tex], we'll use trigonometry. The distance from [tex]\( X \)[/tex] to [tex]\( Y \)[/tex] is 200 km, and the angle from the north is [tex]\( 45^\circ \)[/tex].
- [tex]\( x_Y \)[/tex] (horizontal distance) = [tex]\( 200 \times \cos(45^\circ) \)[/tex]
- [tex]\( y_Y \)[/tex] (vertical distance) = [tex]\( 200 \times \sin(45^\circ) \)[/tex]
### Coordinates Calculation
Using the trigonometric values:
- [tex]\( \cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707 \)[/tex]
Therefore:
- [tex]\( x_Y \approx 200 \times 0.707 = 141.421 \)[/tex] km
- [tex]\( y_Y \approx 200 \times 0.707 = 141.421 \)[/tex] km
So, the coordinates of [tex]\( Y \)[/tex] are [tex]\((141.421, 141.421)\)[/tex].
### Coordinates of [tex]\( Z \)[/tex]
Since [tex]\( Z \)[/tex] is directly east of [tex]\( X \)[/tex], its y-coordinate is the same as that of [tex]\( X \)[/tex], which is 0. Thus, coordinates of [tex]\( Z \)[/tex] can be represented as [tex]\((x_Z, 0)\)[/tex].
### Distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex]
The distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex] is just the horizontal distance [tex]\( x_Z \)[/tex].
Given that [tex]\( x_Y = 141.421 \)[/tex] km and it represents the horizontal distance, we get:
- Distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex] = 141.421 km
### Distance from [tex]\( Y \)[/tex] to [tex]\( Z \)[/tex]
To find the distance from [tex]\( Y \)[/tex] to [tex]\( Z \)[/tex], we'll use the distance formula:
[tex]\[ \text{Distance from } Y \text{ to } Z = \sqrt{(x_Y - x_Z)^2 + (y_Y - y_Z)^2} \][/tex]
Substituting the values we calculated:
- [tex]\( x_Y = 141.421 \)[/tex]
- [tex]\( x_Z = 141.421 \)[/tex]
- [tex]\( y_Y = 141.421 \)[/tex]
- [tex]\( y_Z = 0 \)[/tex]
[tex]\[ \text{Distance from } Y \text{ to } Z = \sqrt{(141.421 - 141.421)^2 + (141.421 - 0)^2} = \sqrt{0 + 141.421^2} = 141.421 \text{ km} \][/tex]
### Summary of Results
Therefore, the distances are:
- (a) The distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex] is 141.421 km.
- (b) The distance from [tex]\( Y \)[/tex] to [tex]\( Z \)[/tex] is also 141.421 km.
Hence, both distances are 141.421 km, correct to three significant figures.
### Setting Up the Problem
Given:
- The aeroplane flies from town [tex]\( X \)[/tex] to town [tex]\( Y \)[/tex] on a bearing of [tex]\( N 45^\circ E \)[/tex].
- Distance from [tex]\( X \)[/tex] to [tex]\( Y \)[/tex] is 200 km.
- Town [tex]\( Z \)[/tex] is directly east of town [tex]\( X \)[/tex].
- Y is on a line with bearing [tex]\( N 45^\circ E \)[/tex] from X, which means a 45-degree angle from the north towards the east.
First, we'll determine the coordinates of [tex]\( Y \)[/tex] relative to [tex]\( X \)[/tex]. Let's place [tex]\( X \)[/tex] at the origin [tex]\((0, 0)\)[/tex].
### Coordinates of [tex]\( Y \)[/tex]
To find the coordinates of [tex]\( Y \)[/tex], we'll use trigonometry. The distance from [tex]\( X \)[/tex] to [tex]\( Y \)[/tex] is 200 km, and the angle from the north is [tex]\( 45^\circ \)[/tex].
- [tex]\( x_Y \)[/tex] (horizontal distance) = [tex]\( 200 \times \cos(45^\circ) \)[/tex]
- [tex]\( y_Y \)[/tex] (vertical distance) = [tex]\( 200 \times \sin(45^\circ) \)[/tex]
### Coordinates Calculation
Using the trigonometric values:
- [tex]\( \cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707 \)[/tex]
Therefore:
- [tex]\( x_Y \approx 200 \times 0.707 = 141.421 \)[/tex] km
- [tex]\( y_Y \approx 200 \times 0.707 = 141.421 \)[/tex] km
So, the coordinates of [tex]\( Y \)[/tex] are [tex]\((141.421, 141.421)\)[/tex].
### Coordinates of [tex]\( Z \)[/tex]
Since [tex]\( Z \)[/tex] is directly east of [tex]\( X \)[/tex], its y-coordinate is the same as that of [tex]\( X \)[/tex], which is 0. Thus, coordinates of [tex]\( Z \)[/tex] can be represented as [tex]\((x_Z, 0)\)[/tex].
### Distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex]
The distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex] is just the horizontal distance [tex]\( x_Z \)[/tex].
Given that [tex]\( x_Y = 141.421 \)[/tex] km and it represents the horizontal distance, we get:
- Distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex] = 141.421 km
### Distance from [tex]\( Y \)[/tex] to [tex]\( Z \)[/tex]
To find the distance from [tex]\( Y \)[/tex] to [tex]\( Z \)[/tex], we'll use the distance formula:
[tex]\[ \text{Distance from } Y \text{ to } Z = \sqrt{(x_Y - x_Z)^2 + (y_Y - y_Z)^2} \][/tex]
Substituting the values we calculated:
- [tex]\( x_Y = 141.421 \)[/tex]
- [tex]\( x_Z = 141.421 \)[/tex]
- [tex]\( y_Y = 141.421 \)[/tex]
- [tex]\( y_Z = 0 \)[/tex]
[tex]\[ \text{Distance from } Y \text{ to } Z = \sqrt{(141.421 - 141.421)^2 + (141.421 - 0)^2} = \sqrt{0 + 141.421^2} = 141.421 \text{ km} \][/tex]
### Summary of Results
Therefore, the distances are:
- (a) The distance from [tex]\( X \)[/tex] to [tex]\( Z \)[/tex] is 141.421 km.
- (b) The distance from [tex]\( Y \)[/tex] to [tex]\( Z \)[/tex] is also 141.421 km.
Hence, both distances are 141.421 km, correct to three significant figures.
