An observer (O) spots a plane flying at a 42° angle to his horizontal line of sight. If the plane is flying at an altitude of 15,000 feet, what is the distance (x) from the plane (P) to the observer (O)?
A right triangle is shown with angle O marked 42 degrees, hypotenuse marked x, and height marked 15,000 feet.
10,035 feet
16,648 feet
20,188 feet
22,417 feet
Answers
Answer:
22,417
Step-by-step explanation:
Just took the test.
Pythagoras theorem states that for a right angle triangle the square of the hypotenuse is equal to the sum of the other two sides of the plane. The distance (x) from the plane (P) to the observer (O) is 22417.75 feet.
Given information-The observer angle to the plane is 42 degrees.
The height of the plane is 15000 feet.
As angle O is 42 degree and the height is 15000. Let suppose the base is y feet long. Thus angle O can be given as,
[tex]\begin{aligned} \tan42^o&=\dfrac{15000}{y} \\ y&=\dfrac{15000}{\tan42^o} \\ y&=16660\\ \end[/tex]
Thus the base of the right triangle is 16660 feet.
Pythagoras theoremPythagoras theorem states that for a right angle triangle the square of the hypotenuse is equal to the sum of the other two sides of the plane.
For the given problem the the hypotenuse side is x feet.
[tex]\begin{aligned}\\ x^2&=15000^2+16660^2\\ x&=22417.75\\ \end[/tex]
Thus the distance (x) from the plane (P) to the observer (O) is 22417.75 feet.
Learn more about the Pythagoras theorem here;
https://brainly.com/question/343682
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I hope I helped!!
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Hope it helps you (^_^)
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