Trigonometry

Mighty Midget · 26-07-2008, 09:31 PM

Here is a problem I really would like to see the answer to, if there is one.

From point A three lines, b, c and d are drawn.

The angles between the three lines are known values

A 4th line intersects these 3 lines at point B, C and D.

The segments BC = CD but have no known value.

Angle ACB + ACD = 180 degrees (due to the fact that the 4th line is a straight line).

The question:

Can one calculate, based on the angles BAC and CAD, the angles ABC, ACB, ACD or ADC?

If so, how?

EDIT: Lately I have started to suspect that it's not the angles BAC and CAD that are important, but rather the relationship between those two angles.

Eagle of Fire · 26-07-2008, 09:49 PM

That's some kind of problem I used to solve in high school...

However, it been so long I have no idea how I did it back in the day. That's the kind of problem you'll never see a viable application IRL unless you're an engineer or something...

Mighty Midget · 26-07-2008, 09:52 PM

The problem was, in fact, found in a computer game where I needed to get a ship's course, preferably from no other data than it's bearings at different time intervals.

The "IRL" problem was: At time t = 0 I had a ship at bearing x. 3 minutes later the bearing had changed to y, and 3 minutes later the bearing was z. From these data I wondered if I could calculate the ship's course.

**_r.u.s.s.** · 26-07-2008, 10:08 PM

i wrote it on paper and scanned, it'd be hell to draw it in mspaint

Mighty Midget · 26-07-2008, 10:10 PM

Sorry, but no. You assumed that the angle ACD was 90 degrees, but that would only be in one particular case, not generally. You can not assume a value for any of the unknown angles.

Also, angle BAC does not necessarily equal angle CAD.

**_r.u.s.s.** · 26-07-2008, 10:12 PM

you said

Quote:

The segments BC = CD but have no known value.

Angle ACB + ACD = 180 degrees (due to the fact that the 4th line is a straight line).

that makes ACD 90 degrees

oops edit: actually i thought you mean something else with second line, nevermind

Mighty Midget · 26-07-2008, 10:13 PM

No, one could be 30 and the other 150 degrees. As for BC = CD, that doesn't guaranty a right angle either

*Luchsen* · 27-07-2008, 12:23 PM

Quote:

Originally Posted by Mighty Midget

Can one calculate, based on the angles BAC and CAD, the angles ABC, ACB, ACD or ADC?

I would say no, but don't pin me down.

Mighty Midget · 27-07-2008, 12:33 PM

That might very well be possible, seeing the lack of data for this particular psoblem. The interesting bit is to see how any difference between the two known angles influence the direction of the 4th line.
If they are both equal and do not equal to zero, then the 4th line is perpendicular to line c.

If they are both equal and equal to zero, then the 4th line is at 0 or 180 degrees on the c line.

If the angle BAC is greater than CAD, then the 4th line is at an angle "a" on the c line 90 < a < 180

If the angle BAC is smaller than CAD, then the 4th line is at an angle at the c line at 0 < a < 90

So, there is some sort of relationship here. The question is is it predictable and can it be calculated?

**_r.u.s.s.** · 27-07-2008, 12:55 PM

what you want to know are the angles only, right? and if you drag the E line further or closer it doesn't matter for angles, they stay the same, as long as BC=CD of course.

so you can pick any value for BC or CD. then it could be calculated