To solve this problem, we need to determine two key distances:
(a) The distance from [tex]\(X\)[/tex] to [tex]\(Z\)[/tex] (which we will call [tex]\(XZ\)[/tex]).
(b) The distance from [tex]\(Y\)[/tex] to [tex]\(Z\)[/tex] (which we will call [tex]\(YZ\)[/tex]).
Let's start solving step-by-step:
### Step 1: Understanding Bearings and Coordinates
1. Bearing from [tex]\(X\)[/tex] to [tex]\(Y\)[/tex]:
- Bearing [tex]\(N 45^\circ E\)[/tex] means the direction is [tex]\(45^\circ\)[/tex] from the north towards the east.
- This implies [tex]\(XY\)[/tex] forms an angle of [tex]\(45^\circ\)[/tex] with the positive x-axis (east direction).
2. Bearing from [tex]\(Y\)[/tex] to [tex]\(Z\)[/tex]:
- Bearing [tex]\(S 60^\circ E\)[/tex] means the direction is [tex]\(60^\circ\)[/tex] from the south towards the east.
- If we convert this bearing for trigonometric purposes, it can be seen as [tex]\(30^\circ\)[/tex] above the negative y-axis (south direction).
### Step 2: Convert Bearings to Cartesian Coordinates
Let’s define our coordinates:
- Let [tex]\(X\)[/tex] be the origin point: [tex]\( (0, 0) \)[/tex]
- Let [tex]\(Y\)[/tex] be point [tex]\(Y(x, y)\)[/tex]
Using the [tex]\(45^\circ\)[/tex] for [tex]\(XY:\)[/tex]
- [tex]\(Y_x = 200 \cos(45^\circ) = 200 \cdot \frac{\sqrt{2}}{2} = 100\sqrt{2}\)[/tex]
- [tex]\(Y_y = 200 \sin(45^\circ) = 200 \cdot \frac{\sqrt{2}}{2} = 100\sqrt{2}\)[/tex]
So, coordinates of [tex]\(Y\)[/tex] are approximately [tex]\( (141.42, 141.42) \)[/tex].
### Step 3: Calculating [tex]\(XZ\)[/tex]
Since [tex]\(Z\)[/tex] is directly east of [tex]\(X\)[/tex], it lies on the x-axis:
- We know that [tex]\(Z\)[/tex]’s x-coordinate must be determined by considering the horizontal distance traversed.
To determine this, consider the bearings and form an understanding:
- Distance from [tex]\(Y\)[/tex] to the east component due to [tex]\(S 60^\circ E\)[/tex] will be along the x-axis from [tex]\(Y\)[/tex].
Utilizing trigonometric relationships:
- [tex]\(Y_y / \tan(60^\circ)\)[/tex] determines the x-direction displacement from [tex]\(Y\)[/tex]
- So, [tex]\(Z_x from Y = 100\sqrt{2} / \sqrt{3} = \frac{100\sqrt{2}}{\sqrt{3}} \approx 81.65\)[/tex]
Therefore:
[tex]\[ Z_x = Y_x + Z_x from Y = 100\sqrt{2} + \frac{100\sqrt{2}}{\sqrt{3}} = 223.07 \][/tex]
Hence, [tex]\(XZ\)[/tex] distance is approximately [tex]\( 223.071 \)[/tex] km.
### Step 4: Calculating [tex]\(YZ\)[/tex]
Using sine for [tex]\(S 60^\circ E\)[/tex]:
- [tex]\(YZ\)[/tex] is calculated by understanding [tex]\( Y_y\ / \sin(60^\circ)\)[/tex]
[tex]\[ Y_y / \sin(60^\circ) = \( 100\sqrt{2} / (\sqrt(3)/2) = 163.299 \][/tex]
Therefore:
[tex]\[ YZ = 163.299 \ km \][/tex]
### Final Results
(a) The distance from [tex]\(X\)[/tex] to [tex]\(Z\)[/tex] is approximately [tex]\(223.071\)[/tex] km.
(b) The distance from [tex]\(Y\)[/tex] to [tex]\(Z\)[/tex] is approximately [tex]\(163.299\)[/tex] km.
