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Sagot :
Certainly! Let's break down this problem step by step.
### Part (a)
We need to show that [tex]\( XZ = x \sqrt{2(1 - \cos \theta)} \)[/tex].
Given:
- [tex]\(XY = x\)[/tex]
- [tex]\(\angle XYZ = \theta\)[/tex]
Since we have points [tex]\(X\)[/tex], [tex]\(Y\)[/tex], and [tex]\(Z\)[/tex], with [tex]\(Y\)[/tex] equidistant from [tex]\(X\)[/tex] and [tex]\(Z\)[/tex], we can use the Law of Cosines to find the distance [tex]\(XZ\)[/tex].
The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cos(\gamma) \][/tex]
For our triangle [tex]\(XYZ\)[/tex]:
- [tex]\(a = b = XY = x\)[/tex]
- [tex]\(\gamma = \angle XYZ = \theta\)[/tex]
- [tex]\(c = XZ\)[/tex]
Applying these to the Law of Cosines:
[tex]\[ XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\theta) \][/tex]
Since [tex]\(XY = YZ = x\)[/tex]:
[tex]\[ XZ^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\theta) \][/tex]
[tex]\[ XZ^2 = 2x^2 - 2x^2 \cdot \cos(\theta) \][/tex]
[tex]\[ XZ^2 = 2x^2 (1 - \cos(\theta)) \][/tex]
Taking the square root of both sides:
[tex]\[ XZ = \sqrt{2x^2 (1 - \cos(\theta))} \][/tex]
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
Thus, we have shown that:
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
### Part (b)
Now, we need to calculate [tex]\(XZ\)[/tex] to the nearest kilometre given that:
- [tex]\(x = 240 \, \text{km}\)[/tex]
- [tex]\(\theta = 132^\circ\)[/tex]
Using the formula derived in part (a):
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
First, convert [tex]\(\theta\)[/tex] from degrees to radians because trigonometric functions typically use radians.
[tex]\[ \theta = 132^\circ \][/tex]
[tex]\[ \theta_{\text{radians}} = \frac{132 \times \pi}{180} \approx 2.303 \, \text{radians} \][/tex]
Now, compute [tex]\(\cos(132^\circ)\)[/tex]:
[tex]\[ \cos(132^\circ) \approx -0.6691 \][/tex]
Substitute the values into the formula:
[tex]\[ XZ = 240 \sqrt{2(1 - (-0.6691))} \][/tex]
[tex]\[ XZ = 240 \sqrt{2(1 + 0.6691)} \][/tex]
[tex]\[ XZ = 240 \sqrt{2(1.6691)} \][/tex]
[tex]\[ XZ = 240 \sqrt{3.3382} \][/tex]
[tex]\[ XZ \approx 240 \times 1.832 \][/tex]
[tex]\[ XZ \approx 439 \, \text{km} \][/tex]
Hence, the distance [tex]\(XZ\)[/tex] is approximately 439 kilometres to the nearest kilometre.
### Part (a)
We need to show that [tex]\( XZ = x \sqrt{2(1 - \cos \theta)} \)[/tex].
Given:
- [tex]\(XY = x\)[/tex]
- [tex]\(\angle XYZ = \theta\)[/tex]
Since we have points [tex]\(X\)[/tex], [tex]\(Y\)[/tex], and [tex]\(Z\)[/tex], with [tex]\(Y\)[/tex] equidistant from [tex]\(X\)[/tex] and [tex]\(Z\)[/tex], we can use the Law of Cosines to find the distance [tex]\(XZ\)[/tex].
The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cos(\gamma) \][/tex]
For our triangle [tex]\(XYZ\)[/tex]:
- [tex]\(a = b = XY = x\)[/tex]
- [tex]\(\gamma = \angle XYZ = \theta\)[/tex]
- [tex]\(c = XZ\)[/tex]
Applying these to the Law of Cosines:
[tex]\[ XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\theta) \][/tex]
Since [tex]\(XY = YZ = x\)[/tex]:
[tex]\[ XZ^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\theta) \][/tex]
[tex]\[ XZ^2 = 2x^2 - 2x^2 \cdot \cos(\theta) \][/tex]
[tex]\[ XZ^2 = 2x^2 (1 - \cos(\theta)) \][/tex]
Taking the square root of both sides:
[tex]\[ XZ = \sqrt{2x^2 (1 - \cos(\theta))} \][/tex]
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
Thus, we have shown that:
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
### Part (b)
Now, we need to calculate [tex]\(XZ\)[/tex] to the nearest kilometre given that:
- [tex]\(x = 240 \, \text{km}\)[/tex]
- [tex]\(\theta = 132^\circ\)[/tex]
Using the formula derived in part (a):
[tex]\[ XZ = x \sqrt{2(1 - \cos(\theta))} \][/tex]
First, convert [tex]\(\theta\)[/tex] from degrees to radians because trigonometric functions typically use radians.
[tex]\[ \theta = 132^\circ \][/tex]
[tex]\[ \theta_{\text{radians}} = \frac{132 \times \pi}{180} \approx 2.303 \, \text{radians} \][/tex]
Now, compute [tex]\(\cos(132^\circ)\)[/tex]:
[tex]\[ \cos(132^\circ) \approx -0.6691 \][/tex]
Substitute the values into the formula:
[tex]\[ XZ = 240 \sqrt{2(1 - (-0.6691))} \][/tex]
[tex]\[ XZ = 240 \sqrt{2(1 + 0.6691)} \][/tex]
[tex]\[ XZ = 240 \sqrt{2(1.6691)} \][/tex]
[tex]\[ XZ = 240 \sqrt{3.3382} \][/tex]
[tex]\[ XZ \approx 240 \times 1.832 \][/tex]
[tex]\[ XZ \approx 439 \, \text{km} \][/tex]
Hence, the distance [tex]\(XZ\)[/tex] is approximately 439 kilometres to the nearest kilometre.
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