Which choice shows the image of triangle QRS after the reflection over the x-axis followed by a rotation of 180° around the orig
in?
1 answer:
Answer:
A
Step-by-step explanation:
We see that triangle QRS is located in Quadrant III. (See attachment for labelling of quadrants)
Reflecting it over the x-axis means that we flip it over the x-axis, or we flip it vertically. Doing so, QRS will end up in Quadrant II.
Now, we need to rotate QRS 180 degrees about the origin. The quadrants are all 90 degrees away from each other, which means that if we rotate QRS about the origin 180 degrees, we'll end up two quadrants away from Quadrant II: we'll end in Quadrant IV.
Look at each of the answer choices. The only choice that features Q'R'S' (the image of QRS) in the fourth quadrant is the first choice, or A. So that's the answer.
<em>~ an aesthetics lover</em>
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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.
--------------------
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.
