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[Mighty Midget]

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]

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]

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.

[GTX2GvO]

Originally Posted by Mighty Midget

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.

We GOT E!!!!

How?? Well Easy.. the full length of e is 6 minutes CONSTANT SPEED
Half an e is 3 minutes CONSTANT SPEED..

So when stated that line c (AC) should also be a known, I can say:
Line c (AC) is 5 minutes Long...

Knowing all angles @ A also you can EASILY calculate ALL remaining angles and Distances in degrees and Minutes of CONSTANT SPEED..

So let's say.. Angle Alfa (BAC) is 18* and Angle Beta (CAD) is 24*.

Now I COULD calculate the rest of it all.....

But not @ this hour... (and not without the proper tools (paper calculator))

So I'll get back on this some other time...
(If it's still an issue that needs to be solved)

[Indignus IV]

sounds right! If you got y/x then just plug that in to my junk and you should get the right answer.
How did you get that AC is 5, though?

[DoomYoshi]

Originally Posted by Mighty Midget

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.

Did the ship maintain the same speed? Does the game provide velocity? How do you know the value of the angles?

[AlumiuN]

Gaaah? Why bring up trig? *calms down a little* Sorry, I had a Maths common test this morning. I've done enough differentiation for one day.

[Mighty Midget]

Sorry for the late reply.

Doom: The angles A and B on the last diagram are calculated from 3 readings of the bearing to the ship. A and B are the difference between the 3 readings.

Also, the speed is assumed to be constant, as is the heading.

[_r.u.s.s.]

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

[Mighty Midget]

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.