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[Lulu_Jane]

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

[Mighty Midget]

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.

[_r.u.s.s.]

Originally Posted by Luchsen

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!"

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

[Indignus IV]

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:

[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?

[GTX2GvO]

Originally Posted by Mighty Midget

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.

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..

[Indignus IV]

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.

[Mighty Midget]

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...

[Indignus IV]

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: