a landing airplane, moving in a straight line , passes over two points (p and q) which are 300 feet apart. when the airplane is above p, as seen from Q, its angle of elevation is 60 degress, and when above Q, as seen from p, its angle of elevation is 40 degress. at what distance from q will the airplane land?

To solve this problem, we will use trigonometry and a bit of algebra.

Let's define some variables:
- Let h be the height of the airplane above the ground at the time it passes over points P and Q.
- Let x be the distance from point Q to the point of landing.

Now, let's analyze the situation from point P:
When the airplane is above point P, it forms a right triangle with the ground and the line of sight from Q to the airplane. The angle of elevation is given as 60 degrees. In this triangle, we have the following relationships:
- The opposite side is h (height of the airplane).
- The adjacent side is 300 feet (distance between P and Q).
- The angle opposite to the side h is 60 degrees.

Using trigonometry, we can determine that h/300 = tan(60 degrees) = √3.

Now, let's analyze the situation from point Q:
When the airplane is above point Q, it forms a right triangle with the ground and the line of sight from P to the airplane. The angle of elevation is given as 40 degrees. In this triangle, we have the following relationships:
- The opposite side is h (height of the airplane).
- The adjacent side is x (distance from Q to the point of landing).
- The angle opposite to the side h is 40 degrees.

Using trigonometry, we can determine that h/x = tan(40 degrees).

Since we have the value of h/300 from the first triangle, we can substitute it into the second equation: √3/300 = tan(40 degrees).

To solve for x, we rearrange the equation:
x = h / (tan(40 degrees) * √3/300).

Substituting the value of h/300 = √3 into the equation:
x = (√3 * 300) / (tan(40 degrees) * √3/300).
x = 900 / (tan(40 degrees) * √3/300).

Evaluating this expression will give you the distance from point Q where the airplane will land.