Trigonometry
 

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

http://i189.photobucket.com/albums/z...ho/diagram.jpg

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.

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...

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.

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

http://img145.imageshack.us/img145/27/scan002ln5.th.jpg

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.

you said
that makes ACD 90 degrees

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

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

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

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?

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

But e has no determined direction.

Perhaps _r.u.s.s. is onto something. Like he says, the lengths AB, AC or AD doesn't change the angles I'm after, but once you fix one length, the others are set, even though they are still unknown.

luchsen it has, because there is only one possible way to place the line so the BC and DC are equal

For starter this is how I look @ trigonometry:

http://img529.imageshack.us/img529/1...etryio5.th.png

Therefor your problem will look like the following in my opinion:

http://img516.imageshack.us/img516/8...try2pt3.th.png

Now you end up wit 3 triangles.. (yes 3)

Namely: ABC ; ACD & ABD

To solve your issue I NEED (besides the known 3 angles @ A)
Either 2 sides OR 1 additional angle AND 1 side.

Or else there can't be a solution.

Let's see:

sin BAC / sin CAD = sin ABC / sin ADC

BAC and CAD are known and ABC + ADC = 180 - BAD
At least there is a relationship here.

EDIT: By giving AC a value, you get one side that two triangles share.

Hmmm... That gives me line c.

That means I STILL need 1 of the following:
gamma(bottom), e1, e2, zèta(top)

Cause to solve this I need to start with. (from the perspective of 1 triangle)
2 sides and 1 angle OR
2 angles and 1 side.

We had 1 angle ( alfa, beta & alfa+beta )
And now we got 1 side ( c )

So I still need either one more side OR one more angle.

i think you are looking a bit narrowish on the problem gtx

Have fun fiddling around with that! My math teacher always yelled at me: "You are supposed to think and calculate, and not to insert values randomly!" Indignant me: "I don't do it randomly!"

E.T. would know the answer .... E.T. would .....
http://www.opensmiley.info/smiley/classique/0295.gif

What game is this MM, and did you sell your soul to play it? :)

Hehe nah, sod the game, but it annoys the hell out of me that I can't find any answer when there are all these suggestions there can be one... Silly, I know, but it's really annoying.

you don't need to know it's value, it's enough to think of it as known variable :p

I know this thread is a tad overcooked already, but I noticed it and thought I'd add my two cents.

I attached a scan of my take on it (assigned my own values to simplify it), and underlined the angle answers in pink.

The short and skinny of it is, I'm not sure it is possible because you need to know the ratio between the line seperating the two triangles, and half of line E. Unless you can figure this out (and I'm quite sure you can't) you can't figure out the exact angles, but you find them in terms of variables using the Law of Sines.

Hope that satisfies :amused:

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)

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?

By this piece of quote...

I simply DEMANDED for more non-variables..

The choice of 5 is just something I personally came up with to fill up the actual value.. (Just as the two angles)

It's just that I overlooked the whole e1=e2=3min IRL text of MM..

Or else I had filled this thread with ALL the values several weeks ago already..

ehh - I think Mighty Midget's suggestion that AC can be given an arbitrary value is wrong. Although the two bottom angles are known, they are not arbitrary, per se, but are "known" and therefore AC cannot be given an arbitrary value. The answer would change if AC changed. The point, I think, is to figure out the four top angles in terms of what you know. In this case, we don't know much about the drawing.

AC can be given an arbitrary value, since all that changes is the size of the greater triangle ABD. The different triangles ABD will still have the same shape... I think...

well, AC can be given an arbitrary value as long as line E increases in direct proportion to it. I don't really care what its value is, its just that we need to know proportion y/x to solve the other angles. Its not enough that we know the bottom angles, or that we could give AC an arbitrary value. We need to know the actual proportion of side AC to side E. Without that, I'm fairly convinced that the problem is impossible.

Don't quote me on it, though. :confused:

If

angle ABC = a (alpha)

K = sin BAC / sin CAD and
L = 180 - (BAC + CAD)

then

K = sin a / sin (L - a)

the solution will be at

sin a / sin (L - a) - K = 0

(tested and verified* with a graphical calculator)

* verified as in I didn't get any wrong answers with several different, arbitrary values for BAC and CAD.

Now is the problem, I am unable to isolate alpha in sin (L - a)
If you can do that, you have the formula, I believe.

Well ACTUALLY...

As long as the Angles @ A (or just one of) are known and E is a known value.. (Which is in the IRL issue) I just need the KNOWN value of either another angle or another side (c {AC})

Cause with the Law of sines and the 180 degrees rule I can then calculate ALL the other sides/angles..
(If I'm in the mood for it AND have my equipment @ hand)

The segments at E are, unfortunately, unknown. The only known values are the two angles at A (angle BAC and CAD).

EDIT: Summary

When the function f(a) = sin a / sin (L - a) - K = 0
that's the correct a (angle ABC)

Sorry for the double, just want to keep things tidy in a messy topic

http://i189.photobucket.com/albums/z...ho/trekant.jpg

x / sin A = y / sin a

x = y sin A / sin a

----

x / sin B = y / sin b

x = y sin B / sin b

----

since x = x

sin A / sin a = sin B / sin b

sin A / sin B = sin a / sin b

sin A / sin B = K

b = 180 - (A + B) - a

180 - (A + B) = L

K = sin a / sin (L - a)

sin a / sin (L - a) - K = 0

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

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. :wacko:

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.

Ohh God, schools only 1 week, and i have to repeat all this shit!

I didn't read through the whole topic... You just have to state the simple relationships (equations 1 and 2) between the three angles formed by e with b, c and d (gamma, delta and epsilon in my notation). Then you use the BC=CD equation and presto, I think using the law of sines is the simplest option.

http://s102.photobucket.com/albums/m...th_bearing.png

Thanks a bunch, Japofran! I will look into this, but it looks promising.

Ok, here's another one:

From a defined point on Earth, you know your distance from that point as well as your bearing as seen from that point (that is, if you are due east of that point, you are at bearing 90 degrees, if you are due north, your bearing is 360, and so on). If your coordinates are 136 70, you are at bearing 136 (South East) at a distance of 70 (be it mm, metres, miles or whatever) from the point.

Your coordinates are [b d] (b being bearing, d being distance)

You also know your own heading h, that is which direction you're facing i.e. 25 degrees (North North East).

You are then to move from coordinates [b d] to a new location given by the coordinates [B D]

What is your new heading H, how many degrees will you have to turn and in which direction (smallest angle), and what is the distance from [b d] to [B D]?

Damn you not another one! :D

OK I'm not sure about what you mean. What I've understood is that from your old position you are to steer to a new heading, then advance in a straight line along it. And what you want to know is that heading and the distance you are to advance. Am I correct?

Hehe yeah, another one.

Yes, from the old position located at [e d] you are to turn to a new heading and move along a straight line to a new location [B D].

The question is what will the new heading be, how many degrees to port or starboard is that new heading from your current one, and how far do you need to travel.

IRL: In military aviation, positions are given by comparing your position to a known, fixed position (the Point in this problem). Your relative position is given as two numbers, one for the bearing seen from the fixed location, the other for the distance in nautical miles from that same location. You know your position and you are told there is an unknown aircraft at a different location and you are ordered to intercept. You now need to know which new heading you need to take, the distance to target as well as which direction will give the smallest turn angle (say, for instance, your current heading is 360 and your new heading is 270, then a 90 degrees turn to port is quicker than a 270 degrees turn to starboard).

The easiest way to solve one of these problems is expressing each position as a vector. Then getting the bandit's position relative to yourself is as easy as substracting your position relative to the base from the bandit's position relative to the base. Adding vectors is graphically equivalent to linking them one after the other. (Substracting a vector is merely adding its opposite, for example -OA = AO.)

Adding or substracting vectors is as easy as adding or substracting all their cartesian coordinates respectively. Since you have both vectors in polar coordinates (distance and bearing), you need to convert them from polar to cartesian, then substract them, and then you'll probably want to convert the result back to polar, since it's more useful that way.

These drawings (click to enlarge) show this, point O being the base, A your craft, and B the bandit. The one to the left shows that the bandit's position relative to you (AB) equals the difference between the bandit's position relative to the base (OB) and your position relative to the base (OA). To the right you can see the relationships between polar (distance, bearing) and cartesian (x, y) coordinates for any of the vectors. (Here the bearing is measured clockwise from the north, although the usual convention in geometry is measuring angles anti-clockwise from the horizontal axis, and the formulae would be changed.)

http://s102.photobucket.com/albums/m..._polardiff.png

Many hand calculators will do all this seamlessly, you just input both vectors in polar coordinates and substract them and the conversions will be made internally. (Some not so advanced calculators can't handle vectors of arbitrary dimension, but some can handle complex numbers and these can be equivalent to 2-D vectors.)

You could easily program all the calculations into a computer program. Especially since you want to get automatically even whether to turn to port or starboard. For example a routine to calculate it could be something like this (in pseudocode, it's not in any specific language), being h1 your current heading and h2 the one you must take (that is the bearing of vector AB, of the bandit relative to you):

The abs() function is supposed to return the absolute value of the input. EXIT means to abort the routine.

Anyway I think an expericed pilot would know by himself which way to turn the moment he learns the heading.

True, when you get the new heading, turning isn't hard to figure out, but I'm more interested in the maths behind it.

Another thing is, learning this positioning system takes a while when you're used to think in terms of x/y axis.

As for the distance d, one equation is d = sq.r. (a^2 + b^2 - 2ab cos(alpha-beta)) where
a and b are the distance element in the two coordinates and alpha and beta are the two bearing elements.

No matter which way you go, the calculations are equivalent:

http://s102.photobucket.com/albums/m...polardiff2.png

Otherwise maths would lie. :P Same goes for the angle.