Question

24. A satellite orbiting the earth passes directly overheao at observation stations in Bengaluru
and Shriharikota. At an instant when satellite is between these two stations its angle of
elevation is simultaneously observed to be complernentary angles. If distance between
Bengaluru and Shriharikota is 350 km. Find
Plz give the answer guys
a)
Value of (cot + tane) in terms of altitude of
satellite from earth.
b) If angle of elevations are equal complementary
angles. Find altitude of satellite.
165

Answer 1

Let A = the satellite.
let B = Phoenix
let C = Los Angeles.
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BC = 340 miles equals the distance between Phoenix and LA.
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Drop a perpendicular from A to intersect with BC at point D.
This forms the line AD which is perpendicular to BC at D.
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You have 2 right triangles.
They are:
ABD and ACD
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let BD = x
let DC = y
you have:
x + y = 340 miles
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let AD = z
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z is the perpendicular from the satellite to the ground (line AD intersecting with line BC at point D).
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x is the distance from Phoenix to point D.
y is the distance from Los Angeles to point D.
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Since tangent is equal to opposite divided by adjacent, we have:
tan(B) = z/x
tan(C) = z/y
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This results in z = x*tan(B) and z = y*tan(C)
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Since they both equal to z, then they are both equal to each other, so:
x*tan(B) = y*tan(C)
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Since we know that x+y = 340, then y = 340-x
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We substitute in the equation we just created to get:
x*tan(B) = (340-x)*tan(C)
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We remove parentheses to get:
x*tan(B) = 340*tan(C) - x*tan(C)
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we add x*tan(C) to both sides of this equation to get:
x*tan(B) + x*tan(C) = 340*tan(C)
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we factor x on the left side of this equation to get:
x*(tan(B)+tan(C)) = 340*tan(C)
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We divide both sides of this equation by (tan(B)+tan(C)) to get:
x = 340*tan(C)/(tan(B)+tan(C))
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Since C = 70 degrees and B = 61 degrees, then this equation becomes:
x = 340*tan(70)/(tan(61)+tan(70))


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We use our calculator to get:
x = 205.2372088
This make y = 340 - x to get:
y = 134.7627912
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Since tan(B) = z/x, then z = x*tan(B)
Since B = 61 degrees and x = 205.2372088, then:
z = x*tan(61)
This calculates out to be:
z = 370.2577258
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Since tan(C) = z/y, then z = y*tan(C)
Since C = 70 degrees and y = 134.7627912, then:
z = y*tan(70)
This calculates out to be:
z = 370.2577258
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z calculated both ways is identical proving that our values for x and y are good.
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We're still not done though.
We need the distance from the Satellite to Phoenix.
That would be the hypotenuse of triangle ABD which is the line AB.
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let m = line AB.
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Sin(B) = opposite/hypotenuse = z/m
Cos(B) = adjacent/hypotenuse = x/m
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Either one will get us m.
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Using Sin(B) = z/m, we multiply both sides of this equation by m to get:
m*Sin(B) = z
We divide both sides of this equation by Sin(B) to get:
m = z/Sin(B)
Since we know the value of z and Sin(B), this calculates out to be:
m = 423.335677 miles
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Using Cos(B) = x/m, we multiply both sides of this equation by m to get:
m*Cos(B) = x
We divide both sides of this equation by Cos(B) to get:
m = x/Cos(B)
Since we know the value of x and Cos(B), this calculates out to be:
m = 423.335677 miles.
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The value of m is identical both ways we calculated it which is as it should be so we are good.
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The answer to your question is:
The satellite is 423.335677 miles from Phoenix.