Trig Word Problem: Aeroplane flying at 10000 m is directly overhead.

[Sarie]

I am having trouble drawing a diagram for this problem:
An aeroplane flying at an altitude of 10000 m is directly overhead. 2 minutes later it is at an angle of 38° to the horizontal. How fast is the aeroplane flying in km per hour? (I know that the answer is 384 km/hr, but like I said, I am having trouble getting started due to lack of diagram)

 

[Otis]

I am having trouble drawing a diagram for this problem:

An aeroplane flying at an altitude of 10000 m is directly overhead. 2 minutes later it is at an angle of 38° to the horizontal. How fast is the aeroplane flying in km per hour?

Here ya go.

If you need more help, please post your steps.

 

[Otis]

I ran out of time, last night. I want to explain what some of the wording means, in your exercise. It's a bit sloppy.

An aeroplane flying at an altitude of 10000 m is directly overhead. 2 minutes later it is at an angle of 38° to the horizontal.

Even though they wrote it, they do not mean that the plane makes an angle with the horizontal. The angle is formed between the path of the plane and the observer's line of sight.

I ought to have labeled the vertices of the triangle, in my sketch. Let's call the observer's position on the ground point A. Label that, on your sketch.

At some moment in time, the plane is directly overhead the observer. Let's call the plane's position there point C.

Two minutes later, the plane has moved some distance, to the right. Let's call that last vertex point B. Label points C and B, on your sketch.

Now, the observer's line of sight to the plane at point B is one ray of the 38° angle, and the plane's horizontal path (from C to B) is the other ray.

That's what they meant, when they wrote, "...it is at an angle of 38° to the horizontal...".

Cheers :cool: