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]$\angle X Y Z$[/tex].

a) If [tex]$X Y = x$[/tex] and [tex]$\angle X Y Z = \theta$[/tex], show that [tex]$X Z = x \sqrt{2(1 - \cos \theta)}$[/tex].

b) Calculate [tex]$X Z$[/tex] (to the nearest kilometre) if [tex]$x = 240 \, \text{km}$[/tex] and [tex]$\theta = 132^{\circ}$[/tex].



Answer :

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.