Question

An observer (O) spots a plane flying at a 55° angle to his horizontal line of sight. If the plane is flying at an altitude of 21,000 feet, what is the distance (x) from the plane (P) to the observer (O)? A right triangle is shown with angle O marked 55 degrees, hypotenuse marked x, and height marked 21,000 feet. 20,793 feet 25,636 feet 28,793 feet 36,585 feet.

Answer 1

Angle of elevation is angle measured from horizontal line to line of sight, The distance from the plane to the observer is 25,636 feet approx

What is angle of elevation?

You look straight parallel to ground. But when you have to watch something high, then you take your sight up by moving your head up. The angle from horizontal line to the point where you stopped your head is called angle of elevation. See the figure where angle OPD is the angle when observer saw the plane. It  is measured from the horizontal line OD.

Refer to the attached figure.

It is given that we have

Length of PD = |PD| = 21,000 feet

Angle ∠OPD = 55°

Assuming length of OP = |OP| = x feet, we get the sin ratio as

$sin(55^\circ) = \dfrac{|PD|}{|OP|} = \dfrac{21000}{x}\\\\x = \dfrac{21000}{\sin(55^\circ)} \approx 25636.27 \approx 25636 \: \rm feet$

(tan 55°  was calculated from online calculator)

Thus,

The distance from the plane to the observer is 25,636 feet approx.

Learn more about trigonometric ratios here:
brainly.com/question/22599614