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5. A surveyor is trying to determine the distance between points [tex]\( X \)[/tex] and [tex]\( Z \)[/tex]. However, the distance cannot be determined directly as a ridge lies between the two points. From a point [tex]\( Y \)[/tex], which is equidistant from [tex]\( X \)[/tex] and [tex]\( Z \)[/tex], he measures the angle [tex]\( XY Z \)[/tex].

a) If [tex]\( XY = x \)[/tex] and [tex]\( \angle XYZ = \theta \)[/tex], show that [tex]\( XZ = x \sqrt{2(1 - \cos \theta)} \)[/tex].

b) Calculate [tex]\( XZ \)[/tex] (to the nearest kilometer) if [tex]\( x = 240 \text{ km} \)[/tex] and [tex]\( \theta = 132^{\circ} \)[/tex].

Sagot :

GinnyAnswer
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.
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