Keilantra spots an airplane on radar that is currently approaching in a straight line, and that will fly directly overhead. The plane maintains a constant altitude of 7500 feet. Keilantra initially measures an angle of elevation of 19degrees
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to the plane at point AA. At some later time, she measures an angle of elevation of 34degrees
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to the plane at point BB. Find the distance the plane traveled from point AA to point BB. Round your answer to the nearest foot if necessary.
Let's assume that the distance from point AA to the plane is x feet. Then, the distance from point BB to the plane is x + d feet, where d is the distance traveled by the plane.
Since the angle of elevation at point AA is 19 degrees, we can use the tangent function to find the height of the plane at point A:
tan(19) = height of the plane / x
height of the plane = x * tan(19)
Similarly, the height of the plane at point B can be found using the tangent function and the given angle of elevation of 34 degrees:
tan(34) = height of the plane / (x + d)
height of the plane = (x + d) * tan(34)
Since the plane maintains a constant altitude of 7500 feet, we can set the two expressions for the height of the plane equal to each other and solve for d:
x * tan(19) = (x + d) * tan(34)
x * tan(19) = x * tan(34) + d * tan(34)
d * tan(34) = x * (tan(34) - tan(19))
d = x * (tan(34) - tan(19)) / tan(34)
To find the distance the plane traveled from point AA to point BB, we need to find the value of d. We can do this by substituting the given values into the equation above:
d = x * (tan(34) - tan(19)) / tan(34)
d = x * (0.664 - 0.354) / 0.664
d = x * (0.31) / 0.664
To find the value of x, we can use the tangent function and the given angle of elevation of 19 degrees:
tan(19) = height of the plane / x
x = height of the plane / tan(19)
Given that the altitude of the plane is 7500 feet, we can substitute this value into the equation above:
x = 7500 / tan(19)
x ≈ 23644.5 feet
Substituting this value of x into the equation for d:
d = x * (0.31) / 0.664
d ≈ 23644.5 * 0.31 / 0.664
d ≈ 11099.5 feet
Therefore, the distance the plane traveled from point AA to point BB is approximately 11099.5 feet.