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Sagot :
Sure, let's address each part of the question step-by-step:
### Part (a)
We start by demonstrating that the distance [tex]\( XZ \)[/tex] can be expressed as [tex]\( XZ = x \sqrt{2(1 - \cos \theta)} \)[/tex].
1. Establish the Scenario:
Points [tex]\( X \)[/tex] and [tex]\( Z \)[/tex] are separated by an inaccessible ridge. The surveyor chooses point [tex]\( Y \)[/tex], which is equidistant from both [tex]\( X \)[/tex] and [tex]\( Z \)[/tex]. This creates two sides of a triangle, [tex]\( XY \)[/tex] and [tex]\( YZ \)[/tex], both equal to [tex]\( x \)[/tex].
2. Use of the Law of Cosines:
The law of cosines for triangle [tex]\( XYZ \)[/tex] states:
[tex]\[ XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\theta) \][/tex]
3. Substitute Known Variables:
Since [tex]\( XY = YZ = x \)[/tex]:
[tex]\[ XZ^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\theta) \][/tex]
4. Simplify the Equation:
[tex]\[ XZ^2 = 2x^2 - 2x^2 \cos(\theta) \][/tex]
Factor out [tex]\( 2x^2 \)[/tex] from the right-hand side:
[tex]\[ XZ^2 = 2x^2 \left(1 - \cos(\theta)\right) \][/tex]
5. Taking the Square Root of both sides to solve for [tex]\( XZ \)[/tex]:
[tex]\[ XZ = x \sqrt{2 \left(1 - \cos(\theta)\right)} \][/tex]
Thus, we have shown that [tex]\( XZ = x \sqrt{2(1 - \cos \theta)} \)[/tex].
### Part (b)
Now, let's calculate [tex]\( XZ \)[/tex] given [tex]\( x = 240 \, \text{km} \)[/tex] and [tex]\( \theta = 132^\circ \)[/tex].
1. Convert the Angle to Radians:
We first convert [tex]\( \theta \)[/tex] from degrees to radians since trigonometric functions typically use radians. The angle in radians is:
[tex]\[ \theta_{\text{radians}} = 132^\circ \times \left(\frac{\pi}{180^\circ}\right) \approx 2.3038 \, \text{radians} \][/tex]
2. Calculate [tex]\( \cos(\theta_{\text{radians}}) \)[/tex]:
For [tex]\( \theta = 132^\circ \)[/tex]:
[tex]\[ \cos(\theta_{\text{radians}}) \approx \cos(2.3038) \approx -0.6691 \][/tex]
3. Substitute into the Equation:
Using the expression from part (a):
[tex]\[ XZ = 240 \times \sqrt{2 \left(1 - \cos(132^\circ)\right)} \][/tex]
Substitute [tex]\(\cos(132^\circ)\)[/tex]:
[tex]\[ XZ = 240 \times \sqrt{2 \left(1 - (-0.6691)\right)} \][/tex]
Simplify inside the square root:
[tex]\[ XZ = 240 \times \sqrt{2 \left(1 + 0.6691\right)} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{2 \times 1.6691} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{3.3382} \][/tex]
[tex]\[ XZ \approx 240 \times 1.826 \][/tex]
4. Multiply and Round:
[tex]\[ XZ \approx 438.50181966844843 \, \text{km} \][/tex]
Rounding this result to the nearest kilometre:
[tex]\[ XZ \approx 439 \, \text{km} \][/tex]
Therefore, the distance [tex]\( XZ \)[/tex] to the nearest kilometre is [tex]\( 439 \)[/tex] km.
### Part (a)
We start by demonstrating that the distance [tex]\( XZ \)[/tex] can be expressed as [tex]\( XZ = x \sqrt{2(1 - \cos \theta)} \)[/tex].
1. Establish the Scenario:
Points [tex]\( X \)[/tex] and [tex]\( Z \)[/tex] are separated by an inaccessible ridge. The surveyor chooses point [tex]\( Y \)[/tex], which is equidistant from both [tex]\( X \)[/tex] and [tex]\( Z \)[/tex]. This creates two sides of a triangle, [tex]\( XY \)[/tex] and [tex]\( YZ \)[/tex], both equal to [tex]\( x \)[/tex].
2. Use of the Law of Cosines:
The law of cosines for triangle [tex]\( XYZ \)[/tex] states:
[tex]\[ XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\theta) \][/tex]
3. Substitute Known Variables:
Since [tex]\( XY = YZ = x \)[/tex]:
[tex]\[ XZ^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\theta) \][/tex]
4. Simplify the Equation:
[tex]\[ XZ^2 = 2x^2 - 2x^2 \cos(\theta) \][/tex]
Factor out [tex]\( 2x^2 \)[/tex] from the right-hand side:
[tex]\[ XZ^2 = 2x^2 \left(1 - \cos(\theta)\right) \][/tex]
5. Taking the Square Root of both sides to solve for [tex]\( XZ \)[/tex]:
[tex]\[ XZ = x \sqrt{2 \left(1 - \cos(\theta)\right)} \][/tex]
Thus, we have shown that [tex]\( XZ = x \sqrt{2(1 - \cos \theta)} \)[/tex].
### Part (b)
Now, let's calculate [tex]\( XZ \)[/tex] given [tex]\( x = 240 \, \text{km} \)[/tex] and [tex]\( \theta = 132^\circ \)[/tex].
1. Convert the Angle to Radians:
We first convert [tex]\( \theta \)[/tex] from degrees to radians since trigonometric functions typically use radians. The angle in radians is:
[tex]\[ \theta_{\text{radians}} = 132^\circ \times \left(\frac{\pi}{180^\circ}\right) \approx 2.3038 \, \text{radians} \][/tex]
2. Calculate [tex]\( \cos(\theta_{\text{radians}}) \)[/tex]:
For [tex]\( \theta = 132^\circ \)[/tex]:
[tex]\[ \cos(\theta_{\text{radians}}) \approx \cos(2.3038) \approx -0.6691 \][/tex]
3. Substitute into the Equation:
Using the expression from part (a):
[tex]\[ XZ = 240 \times \sqrt{2 \left(1 - \cos(132^\circ)\right)} \][/tex]
Substitute [tex]\(\cos(132^\circ)\)[/tex]:
[tex]\[ XZ = 240 \times \sqrt{2 \left(1 - (-0.6691)\right)} \][/tex]
Simplify inside the square root:
[tex]\[ XZ = 240 \times \sqrt{2 \left(1 + 0.6691\right)} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{2 \times 1.6691} \][/tex]
[tex]\[ XZ = 240 \times \sqrt{3.3382} \][/tex]
[tex]\[ XZ \approx 240 \times 1.826 \][/tex]
4. Multiply and Round:
[tex]\[ XZ \approx 438.50181966844843 \, \text{km} \][/tex]
Rounding this result to the nearest kilometre:
[tex]\[ XZ \approx 439 \, \text{km} \][/tex]
Therefore, the distance [tex]\( XZ \)[/tex] to the nearest kilometre is [tex]\( 439 \)[/tex] km.
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