# Trig -- An airplane is ascending at an angle of 10 degrees

1. Sep 26, 2014

### Hierophant

1. The problem statement, all variables and given/known data
An airplane is ascending at an angle of 10 degrees is detected 2000m directly above an observer after 20 seconds the angle of elevation to the plane is 48 degrees. How fast is the plane going?

2. Relevant equations

3. The attempt at a solution
I just don't know what to do next, I've tried several additional extrapolated triangles. but to no avail. I'm not sure if what I have up to this point is even right!
My attempt is in the associated image

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Last edited by a moderator: Sep 26, 2014
2. Sep 26, 2014

### pasmith

Neither of your diagrams is correct.

The plane travels along a straight line which is at 10° to the horizontal. Initially it is directly above the observer at a height of 2000m. Twenty seconds later, the angle which the line from the observer to the plane makes with the horizontal is 48°.

Let the speed of the plane be $v\,\mathrm{ms}^{-1}$, so that the distance travelled by the plane (along its line of ascent) in 20 seconds is $20v$ meters.

That gives you two relevant lengths. Can you find a triangle which has those as two of its sides?

3. Sep 28, 2014

### Hierophant

So, here's my latest attempt.

I should note that this is in the challenge section of this very short trig book. I'm not sure if I am taking the right approach, I was wondering if you could guide me a bit more, this is killing me!

I don't think the standard approaches provided by the book apply here, or there's some leap that you have to take first. This is second to last question in the book. All before this I was just blitzing, finishing this 70 page text book in two days.

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4. Sep 29, 2014

### pasmith

Your diagram is correct, and the sine rule is the correct method, but there is sufficient information given for you to determine the angle CAB. Remember that you are asked to find the speed $v$.

5. Sep 30, 2014

### HallsofIvy

Staff Emeritus
10 degrees to the horizontal is 80 degrees to the vertical. Further, an angle of elevation of 40 degrees is 50 degrees to the vertical. Those two angles add to 130 degrees so the third angle in the triangle is 180- 130= 50 degrees. The distance between the 80 and 50 degree angles is 2000m so you can use the sine law to find the other lengths. In particular, the distance the airplane flew in 20 seconds is the side opposite the first 50 degree angle.