A plane is going in circle around an airport the plane takes $3$ minutes to complete one round.
The angle of elevation of the plane from point $p$ on the ground at time $t$ seconds is equal to that at time $(t+30)$ seconds.
At time $(t+x)$ seconds the plane flies vertically above the point $p$.
What is $x$ equal to?
I have tried using basic trig, but that doesn't helps, may be there is some use of circle chord or something.
Here is how to proceed:
Assume, at time t, the plane is at the point A on the circle.
After 30s, the plane arrives at the point B. Because it takes 180s to fly the full 360° circle, the arc A to B spans 60°.
Let C be the point on the circle when the plane is directly above p. Then, AC = BC due to the same elevation angles. So, ∆ABC is an isosceles triangle inscribing the circle.
Thus, from A to C, the plane flies an arc of 150°+60° = 210°, or 7/12 (210°/360°) of a full circle.
Therefore, $x$ = (7/12)$\cdot$180s = 105s.
The other possibility is that the point C lies midway along the path AB, the 60°-arc. In this case, the arc AC spans 60°/2 = 30° and $x$ = 15s.