# Trig homework question (yes, it involves fairly advanced physics)

1. Jan 2, 2006

### nealh149

I'm a junior in an honors trigonometry class, and am also studying some more advanced physics on my own. So i've decided to try to solve the problem accounting for the speed of light. here is the problem.

An airplane is sighted at the same time by two ground observers who are 4 miles apart and in line with the airplane. They report the angles of elevation as 15 and 20 degrees. How high is the airplane?

I've rendered this problem unsolvable, and here is my reasons. Because the plane is sighted by both parties simultaneously, that must mean the line of sight distance in both cases must be equal because the speen of light is finite and constant. Therefore, you have 2 systems with two variables each, both have a varying height of the plane and distance to the point under the plane. The only info you have is one angle in each triangle, and that these triangle's hypotonuses are equal.

Does this make sense to anybody else?

2. Jan 2, 2006

### erok81

Unless I am reading the last part wrong, you don't have two triangles with equal hypotenuses. You end up with one triangle, three different angles, and three differing sides.

Also, I think you can take out the differing times they see the plan due to lights speed. Since they are only 4 miles apart and human, there probably isn't much of a difference when the light gets there.

3. Jan 2, 2006

### nealh149

Thanks for the quick response, but I think you misunderstood my first post, i was probably unclear.

The problem states that they see the airplane at the same time. If this means that if it was the EXACT same time, the line of sight distance must be the exact time. If c remains constant, t remains constant, then d must remain constant.

4. Jan 2, 2006

### erok81

Well, maybe it depends on what class this is for. ;)

You were clear in you first post, just maybe looking too far into the question. Since they are so close the time they both see the plane is so small, there is no way they'd notice a difference. But, factor that in and see how small the difference in time is. You will see, it is barely even there.

Does that make sense? :rofl:

5. Jan 2, 2006

### nealh149

Just to make it clear, i'm doing this for my own further understanding, and am not actually handing in this solution.

But to my understanding, wouldn't the times in this case be the same, because both observers saw the plane at the same time. Since this remains constant, so doesn't the distance.

6. Jan 2, 2006

### HallsofIvy

Staff Emeritus
No, that's wrong. The fact that the two observers see the plane at the same time means that the observers are seeing the plane in two different places in its flight.

If the height of the airplane is "h", you can calculate the lengths of the hypotenuses of the two right triangles as a function of h. Given that, you can find the difference in times for light to travel to each observer. It might well be that you need to know the speed of the airplane in order to answer this.