# A very formulaic trig word problem (find an angle)

1. May 13, 2012

### solve

1. The problem statement, all variables and given/known data

A vacation resort in a mountain town has installed a zip line( a sturdy wire, down which costumers in harnesses can quickly descend from high altitudes) to attract patrons. One zip line is 1,750 feet long and allows its rider to descend from a ski slope down to the ground, a vertical drop of 450 feet. Calculate the angle of declension of the wire in radians, accurate to three decimal places.

2. Relevant equations
3. The attempt at a solution

I have a question about the drawing of this situation. I don't think I can draw it here so hopefully you can see what I am trying to draw from a little algebra work. See, the picture is drawn such that the angle works out to x=arcsin(9/35) from sinx=450/1,750. The top of the mountain is a plane.

I drew it in way that the angle works out to x=arccos(9/35) with adjacent side equaling 450 feet. In other words, my triangle is flipped upside down.

Why am I wrong? Thanks

2. May 13, 2012

### Infinitum

The question asks you to find the angle of declination from the horizontal level.

The hill is something like this

|- -------------
|. -
|... -
|..... -
|_____-

And you need to find the angle between the upper horizontal line and the slanted line. Your answer gives the angle between the slanted line and the vertical one.

3. May 13, 2012

### solve

Oh, I see now. Thanks, Infinitum.

edit:

Let's say there was no picture that I could look up in relation to this situation. Would it be really wrong, then, to assume that there is no horizontal line on top of the mountain, because I really don't get the reference to that from the question? What if that was said explicitly? Would then the angle between the slanted line and the vertical one be considered the angle of declination? Thanks.

Last edited: May 13, 2012
4. May 13, 2012

### Infinitum

The angle of declination(depression) by definition means from a given horizontal level. This diagram should clear it up for you.

5. May 13, 2012

### solve

Ah-huh! So the elevation angle would be the one between the horizontal line (the ground) and the sight line? Acute one? Obtuse one?

Last edited: May 13, 2012
6. May 13, 2012

### Infinitum

Yep. It might change to a different horizontal line depending on the problem, but it usually is the ground.

7. May 13, 2012

### solve

Cool. In the drawing above which angle is the elevation angle? The obtuse one or the acute one?

Thank You.

8. May 13, 2012

### Infinitum

As far as I know, elevation and depression angles are acute. It just doesn't feel right to let them be obtuse

9. May 13, 2012

### solve

That's it, Infinitum. I appreciate your hep and thank you.

10. May 13, 2012

### Infinitum

You're welcome!

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