What's the solution set of –5 < 2x – 1 < 3?
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Answers:
- Satellite is approximately <u>2446.43 km</u> from station A.
- Satellite is approximately <u>2441.61 km</u> above the ground.
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Explanation:
I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.
The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths
- AB = 60 km
- AD = x
- CD = h
Focusing on triangle ACD, we can apply the tangent rule to isolate h.
tan(angle) = opposite/adjacent
tan(A) = CD/AD
tan(86.4) = h/x
x*tan(86.4) = h
h = x*tan(86.4)
We'll use this later in the substitution below.
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Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say
tan(angle) = opposite/adjacent
tan(B) = CD/DB
tan(B) = CD/(DA+AB)
tan(85) = h/(x+60)
tan(85)*(x+60) = h
tan(85)*(x+60) = x*tan(86.4) ............. apply substitution; isolate x
x*tan(85)+60*tan(85) = x*tan(86.4)
60*tan(85) = x*tan(86.4)-x*tan(85)
60*tan(85) = x*(tan(86.4)-tan(85))
x*(tan(86.4)-tan(85)) = 60*tan(85)
x = 60*tan(85)/(tan(86.4)-tan(85))
x = 153.612786190499
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We'll use this approximate x value to find h
h = x*tan(86.4)
h = 153.612786190499*tan(86.4)
h = 2441.60531869599
h = 2441.61 km is how high the satellite is above the ground.
Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC
cos(angle) = adjacent/hypotenuse
cos(A) = AD/AC
cos(86.4) = x/AC
cos(86.4) = 153.612786190499/AC
AC*cos(86.4) = 153.612786190499
AC = 153.612786190499/cos(86.4)
AC = 2446.43279498247
AC = 2446.43 km is the distance from the satellite to station A.
