Mary leaves her house to take a walk. The graph shows the distance, d, in feet from her house that Mary is at any given time, t,
in minutes. How many minutes has Mary walked when she reaches the farthest point from her house?
1 answer:
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However, we can't solve it without any additional information, all we know is that
However, we can't solve it without any additional information, all we know is that
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.
