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Sagot :
Let's tackle the problem presented by the surveyor step-by-step.
## Part (a):
We start with point [tex]\(Y\)[/tex] being equidistant from points [tex]\(X\)[/tex] and [tex]\(Z\)[/tex], therefore [tex]\(XY = YZ = x\)[/tex].
The problem requires us to find [tex]\(XZ\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(\theta\)[/tex].
To solve this, we will use the Law of Cosines, which states:
[tex]\[ c^2 = a^2 + b^2 - 2ab\cos(\theta) \][/tex]
In this scenario:
- [tex]\(a = XY = x\)[/tex]
- [tex]\(b = YZ = x\)[/tex]
- [tex]\(\theta = \angle XYZ\)[/tex]
- [tex]\(c = XZ\)[/tex]
Plugging these values into the Law of Cosines, we get:
[tex]\[ XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\theta) \][/tex]
Substituting [tex]\(XY = x\)[/tex] and [tex]\(YZ = x\)[/tex]:
[tex]\[ XZ^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\theta) \][/tex]
Simplify the equation:
[tex]\[ XZ^2 = 2x^2 - 2x^2 \cos(\theta) \][/tex]
[tex]\[ XZ^2 = 2x^2 (1 - \cos(\theta)) \][/tex]
Taking the square root on both sides to solve for [tex]\(XZ\)[/tex]:
[tex]\[ XZ = \sqrt{2x^2(1 - \cos(\theta))} \][/tex]
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
Hence, we have shown that:
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
## Part (b):
Now, we need to calculate [tex]\(XZ\)[/tex] if [tex]\(x = 240 \, \text{km}\)[/tex] and [tex]\(\theta = 132^\circ\)[/tex].
### Step-by-step Calculation:
1. Convert [tex]\(\theta\)[/tex] from degrees to radians:
[tex]\[ \theta = 132^\circ \][/tex]
The conversion formula is [tex]\( \theta \text{ radians} = \theta \text{ degrees} \times \left(\frac{\pi}{180}\right) \)[/tex]:
[tex]\[ \theta = 132 \times \left(\frac{\pi}{180}\right) \approx 2.303 \text{ radians} \][/tex]
2. Calculate [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos(132^\circ) = \cos(2.303 \text{ radians}) \approx -0.669 \][/tex]
3. Plug these values into the formula [tex]\(XZ = x \sqrt{2(1 - \cos(\theta))}\)[/tex]:
[tex]\[ XZ = 240 \, \text{km} \times \sqrt{2(1 - (-0.669))} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{2(1 + 0.669)} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{2(1.669)} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{3.338} \][/tex]
[tex]\[ XZ \approx 240 \times 1.827 \][/tex]
[tex]\[ XZ \approx 438.48 \][/tex]
4. Rounding to the nearest kilometer:
[tex]\[ XZ \approx 439 \, \text{km} \][/tex]
Therefore, the distance between points [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] is approximately [tex]\(439\)[/tex] kilometers.
## Part (a):
We start with point [tex]\(Y\)[/tex] being equidistant from points [tex]\(X\)[/tex] and [tex]\(Z\)[/tex], therefore [tex]\(XY = YZ = x\)[/tex].
The problem requires us to find [tex]\(XZ\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(\theta\)[/tex].
To solve this, we will use the Law of Cosines, which states:
[tex]\[ c^2 = a^2 + b^2 - 2ab\cos(\theta) \][/tex]
In this scenario:
- [tex]\(a = XY = x\)[/tex]
- [tex]\(b = YZ = x\)[/tex]
- [tex]\(\theta = \angle XYZ\)[/tex]
- [tex]\(c = XZ\)[/tex]
Plugging these values into the Law of Cosines, we get:
[tex]\[ XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\theta) \][/tex]
Substituting [tex]\(XY = x\)[/tex] and [tex]\(YZ = x\)[/tex]:
[tex]\[ XZ^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\theta) \][/tex]
Simplify the equation:
[tex]\[ XZ^2 = 2x^2 - 2x^2 \cos(\theta) \][/tex]
[tex]\[ XZ^2 = 2x^2 (1 - \cos(\theta)) \][/tex]
Taking the square root on both sides to solve for [tex]\(XZ\)[/tex]:
[tex]\[ XZ = \sqrt{2x^2(1 - \cos(\theta))} \][/tex]
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
Hence, we have shown that:
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
## Part (b):
Now, we need to calculate [tex]\(XZ\)[/tex] if [tex]\(x = 240 \, \text{km}\)[/tex] and [tex]\(\theta = 132^\circ\)[/tex].
### Step-by-step Calculation:
1. Convert [tex]\(\theta\)[/tex] from degrees to radians:
[tex]\[ \theta = 132^\circ \][/tex]
The conversion formula is [tex]\( \theta \text{ radians} = \theta \text{ degrees} \times \left(\frac{\pi}{180}\right) \)[/tex]:
[tex]\[ \theta = 132 \times \left(\frac{\pi}{180}\right) \approx 2.303 \text{ radians} \][/tex]
2. Calculate [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos(132^\circ) = \cos(2.303 \text{ radians}) \approx -0.669 \][/tex]
3. Plug these values into the formula [tex]\(XZ = x \sqrt{2(1 - \cos(\theta))}\)[/tex]:
[tex]\[ XZ = 240 \, \text{km} \times \sqrt{2(1 - (-0.669))} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{2(1 + 0.669)} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{2(1.669)} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{3.338} \][/tex]
[tex]\[ XZ \approx 240 \times 1.827 \][/tex]
[tex]\[ XZ \approx 438.48 \][/tex]
4. Rounding to the nearest kilometer:
[tex]\[ XZ \approx 439 \, \text{km} \][/tex]
Therefore, the distance between points [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] is approximately [tex]\(439\)[/tex] kilometers.
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