An aircraft which is maintaining a constant altitude of 1000 metre is travelling directly away from an observer. At a particular instant the angle of elevation is 52 degree and 12 seconds later the angle of elevation is 16 degree. Find the distance travelled in this time.

To find the distance traveled by the aircraft, we can consider the observer, the aircraft, and the ground as forming a right triangle.

Let's assume that the initial position of the aircraft is at point A, and after 12 seconds it is at point B. Point O represents the position of the observer, and C represents the point on the ground directly below point B.

We can now apply trigonometry to solve the problem. Since the aircraft is maintaining a constant altitude of 1000 meters and the angle of elevation decreases from 52 degrees to 16 degrees, we can use the tangent function to relate the angles and the distances involved.

In triangle OAC:
tan(52 degrees) = height (AB) / distance (AC)

In triangle OBC:
tan(16 degrees) = height (BC) / distance (AC + x)

Where x represents the distance traveled by the aircraft in the 12-second time interval.

Now, we can rearrange the equations to isolate the distances:

AB = tan(52 degrees) * AC
BC = tan(16 degrees) * (AC + x)

We can equate AB and BC to find the distance traveled:

tan(52 degrees) * AC = tan(16 degrees) * (AC + x)

To solve for x, we can rearrange the equation:

tan(52 degrees) * AC = tan(16 degrees) * AC + tan(16 degrees) * x

Simplifying:

tan(52 degrees) * AC - tan(16 degrees) * AC = tan(16 degrees) * x

Factoring out AC:

( tan(52 degrees) - tan(16 degrees) ) * AC = tan(16 degrees) * x

Dividing both sides by ( tan(52 degrees) - tan(16 degrees) ):

AC = ( tan(16 degrees) * x ) / ( tan(52 degrees) - tan(16 degrees) )

Now, we can substitute the given values into the equation to find the distance traveled:

AC = ( tan(16 degrees) * 12 ) / ( tan(52 degrees) - tan(16 degrees) )