Question

2. A small plane is at a point P, which is 2600 feet directly above an observer on the ground at a point O. The plane then flies 4500 feet to a point Q, where the line between the observer and the plane makes a 50∘ angle with the vertical. a) What is the distance between the observer and Q ? b) How much elevation did the plane gain between P and Q ?

Answer 1

The distance between the observer and point Q is approximately 3441.40 feet, and the plane gained an elevation of 841.40 feet between points P and Q.

Step-by-step explanation:

To find the distance between the observer and point Q, we can use trigonometry. Since the observer is directly below point P and both Q and the observer form a right-angled triangle with the line between them as the hypotenuse, we can use the sine function to find the length of the adjacent side. Using the given angle of 50∘ and the distance of 4500 feet, we have:

Adjacent side = hypotenuse * sin(angle) = 4500 * sin(50) = 3441.40 feet.

Therefore, the distance between the observer and point Q is approximately 3441.40 feet.

To find the elevation of the plane gained between P and Q, we can use simple subtraction. Since the plane moves directly from P to Q, the elevation gained is the difference between the altitudes of the two points. Given that point P is 2600 feet above the ground and point Q is 3441.40 feet above the ground, we have:

Elevation gained = 3441.40 - 2600 = 841.40 feet.

Answer 2

The student's question involves using trigonometry to find the distance between an observer and a plane's new location, as well as the elevation gained by the plane. The horizontal distance is found using the tangent function and the change in altitude by the sine function and subtracting the initial altitude.

Step-by-step explanation:

The student's question involves determining the distance between an observer and a plane at a different location, and the elevation gained by the plane during its flight from one point to another. To solve this problem, trigonometric principles and right-triangle properties can be used.

Part a: Distance between the observer and point Q

To find the horizontal distance from the observer at point O to point Q, we can use the tangent function. Since the plane makes a 50° angle with the vertical, we can express this as:

OQ = OP * tan(50°)

Where OQ is the horizontal distance from O to Q, and OP is the vertical distance, which is 2600 feet. Plugging in the values:

OQ = 2600 * tan(50°)

After calculating, we find the horizontal distance OQ.

Part b: Elevation gained by the plane

To determine the elevation gained, we find the vertical distance of point Q above the observer minus the original altitude:

Elevation gain = PQ - OP

Where PQ is the hypotenuse (4500 feet), and we can calculate the vertical component (PQ's vertical projection) using:

PQ_vertical = PQ * sin(50°)

Subtracting the original altitude (OP) from this value gives us the elevation gained.