Get personalized answers to your specific questions with IDNLearn.com. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.
Sagot :
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.
(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.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.