Question

1. the diagram illustrates a birds - eye view of two cell towers and a control station. how much closer is tower a to the station than tower b?

Explanation:

Step1: Identify the triangle type

The diagram shows a right - triangle with hypotenuse \( c = 140\space m \) and an angle \( \theta=42^{\circ} \). We need to find the adjacent side (distance from station to Tower A, let's call it \( x \)) and the opposite side? Wait, no. Wait, actually, we can use trigonometric ratios. Let's assume that the distance from Tower B to the station is the hypotenuse? Wait, no, looking at the diagram, the triangle has a right angle at the station, angle at Tower B is \( 42^{\circ} \), and the side opposite to the right angle (hypotenuse) is the distance between Tower A and Tower B? Wait, no, the problem is to find how much closer Tower A is to the station than Tower B. Let's denote:

Let the distance from Tower A to the station be \( d_A \), from Tower B to the station be \( d_B \), and the distance between Tower A and Tower B be \( 140\space m \). The angle at Tower B is \( 42^{\circ} \).

In the right - triangle, \( \cos(42^{\circ})=\frac{d_A}{140} \) (adjacent over hypotenuse) and \( \sin(42^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}} \)? Wait, no, if the right angle is at the station, then:

- The side adjacent to the \( 42^{\circ} \) angle (at Tower B) is the distance from Tower A to the station (\( d_A \)).
- The side opposite to the \( 42^{\circ} \) angle is the distance from Tower B to the station? No, wait, no. Let's correct:

Let’s define:

- Let the station be point \( S \), Tower A be \( A \), Tower B be \( B \). So \( \triangle ASB \) is right - angled at \( S \), \( \angle B = 42^{\circ} \), and \( AB=140\space m \).

We know that in right - triangle \( ASB \):

\( \cos(\angle B)=\frac{SB}{AB} \) (adjacent over hypotenuse) and \( \sin(\angle B)=\frac{AS}{AB} \)? Wait, no:

\( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \), \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \)

Here, \( \theta = 42^{\circ} \), hypotenuse \( AB = 140\space m \), adjacent side to \( \theta \) is \( SB \) (distance from Tower B to station), opposite side is \( AS \) (distance from Tower A to station). Wait, but we need to find \( SB - AS \) (how much closer Tower A is to the station than Tower B? Wait, no, if \( AS \) is the distance from A to S and \( SB \) is from B to S, then the difference \( SB - AS \) is how much closer A is to S than B is to S.

First, find \( AS \) and \( SB \):

\( AS=AB\times\sin(42^{\circ}) \)

\( SB = AB\times\cos(42^{\circ}) \)

Then the difference \( \Delta d=SB - AS=AB(\cos(42^{\circ})-\sin(42^{\circ})) \)

Step2: Calculate the values

We know that \( AB = 140\space m \), \( \cos(42^{\circ})\approx0.7431 \), \( \sin(42^{\circ})\approx0.6691 \)

First, calculate \( SB=140\times\cos(42^{\circ})\approx140\times0.7431 = 104.034\space m \)

Second, calculate \( AS = 140\times\sin(42^{\circ})\approx140\times0.6691=93.674\space m \)

Then the difference \( \Delta d=SB - AS=104.034 - 93.674 = 10.36\space m \) (approximate value). Wait, but maybe we made a mistake in the trigonometric ratio. Wait, maybe the hypotenuse is the distance from Tower B to Tower A, and we need to find the difference in their distances to the station.

Wait, another approach:

Let the distance from Tower A to station be \( x \), from Tower B to station be \( y \). The triangle is right - angled at the station, so by Pythagoras, \( x^{2}+y^{2}=140^{2} \), and \( \tan(42^{\circ})=\frac{x}{y} \) (since \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \), if \( \theta = 42^{\circ} \) at Tower B, then \( \tan(42^{\circ})=\frac{AS}{SB}=\frac{x}{y} \), so \( x = y\tan(42^{\circ}) \)

Substitute into Pythagoras: \( (y\tan(42^{\circ}))^{2}+y^{2}=140^{2} \)

\( y^{2}(\tan^{2}(42^{\circ}) + 1)=140^{2} \)

Since \( \tan^{2}\theta+ 1=\sec^{2}\theta=\frac{1}{\cos^{2}\theta} \)

So \( y^{2}=\frac{140^{2}\cos^{2}(42^{\circ})}{1} \)

\( y = 140\cos(42^{\circ}) \), and \( x = 140\sin(42^{\circ}) \)

Then the difference \( y - x=140(\cos(42^{\circ})-\sin(42^{\circ})) \)

Calculate \( \cos(42^{\circ})\approx0.7431 \), \( \sin(42^{\circ})\approx0.6691 \)

\( \cos(42^{\circ})-\sin(42^{\circ})\approx0.7431 - 0.6691 = 0.074 \)

\( 140\times0.074 = 10.36\space m \) (approximate)

Wait, but maybe the diagram is such that the hypotenuse is the distance from Tower B to Tower A, and we need to find the adjacent and opposite sides. Alternatively, maybe we should use:

If we consider that the distance from Tower A to the station is \( d_A=140\cos(42^{\circ}) \) and from Tower B to the station is \( d_B = 140\sin(42^{\circ}) \)? No, that would be wrong. Wait, let's re - examine the angle.

Wait, the right angle is at the station. So:

- Angle at Tower B: \( 42^{\circ} \)
- Side \( AB = 140\space m \) (hypotenuse)
- Side \( AS \): opposite to angle \( B \), so \( AS = AB\sin(42^{\circ}) \)
- Side \( SB \): adjacent to angle \( B \), so \( SB=AB\cos(42^{\circ}) \)

We need to find how much closer Tower A is to the station than Tower B. So we need to find \( SB - AS \) (because if \( SB>AS \), then A is closer to S than B is, so the difference is \( SB - AS \))

\( SB - AS=AB\cos(42^{\circ})-AB\sin(42^{\circ})=AB(\cos(42^{\circ})-\sin(42^{\circ})) \)

\( AB = 140\space m \)

\( \cos(42^{\circ})\approx0.7431 \), \( \sin(42^{\circ})\approx0.6691 \)

\( \cos(42^{\circ})-\sin(42^{\circ})\approx0.7431 - 0.6691=0.074 \)

\( 140\times0.074 = 10.36\space m \approx10\space m \) (or more accurately, let's use more precise values)

Using calculator - precise values:

\( \cos(42^{\circ})\approx0.7431448255 \)

\( \sin(42^{\circ})\approx0.6691306064 \)

\( \cos(42^{\circ})-\sin(42^{\circ})\approx0.7431448255 - 0.6691306064 = 0.0740142191 \)

\( 140\times0.0740142191\approx10.36\space m \)

Answer:

Approximately \( 10.4\space m \) (or \( 10\space m \) depending on the required precision)