Trigonometry to Find Stronghold in Minecraft

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In Minecraft, players can locate strongholds by throwing "eye of enders," which indicate the direction of the stronghold but are rare to obtain. A user proposed using trigonometry, specifically the law of sines, to determine the coordinates of the stronghold by throwing two eyes and measuring angles. However, it was noted that the eyes do not provide angle measurements, making this method potentially inaccurate. Instead, continuing in the direction of the first eye until reaching the stronghold is recommended. The discussion emphasizes the challenges of triangulating positions accurately in the game.
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I play this game, Minecraft, and I have to find a place called stronghold, which can only be found by throwing "eye of enders" in the air, that go in the direction of where the stronghold is hidden. The only problem is, those eyes are very rare and difficult to get, so I thought something about this: by throwing only 2 ender eyes, I can get the coordinates of the stronghold. The picture might be more understandable. You guys think this can be done by using trigonometry? What formula should I use?? :)

Thanks in advance,
Nexor.

EDIT: x3,y3 would be the stronghold's location. :)
 

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Nexor said:
I play this game, Minecraft, and I have to find a place called stronghold, which can only be found by throwing "eye of enders" in the air, that go in the direction of where the stronghold is hidden. The only problem is, those eyes are very rare and difficult to get, so I thought something about this: by throwing only 2 ender eyes, I can get the coordinates of the stronghold. The picture might be more understandable. You guys think this can be done by using trigonometry? What formula should I use?? :)

Thanks in advance,
Nexor.

EDIT: x3,y3 would be the stronghold's location. :)

If you can think of a way to measure angles using minecraft, it would be possible using the very simple law of sines. But I don't think the eye of enders give you any measures of angles or distances.
 


Why even throw two? Just continue in the direction given by the first until you get to the right place. Yes, if it is invisible or hard to see even when you are on top of it, "triangulating" like that will give the position to look.

Since you know two angles and two angles, you can find the third: 180- \theta_1- \theta_2. And then, knowing one distance (between the two points where you "throw the eyes") you can use the sine law:
\frac{a}{sin(A)}= \frac{b}{sin(B)}= \frac{c}{sin(C)}
to find the other two lengths.

But that is assuming your directions are perfect. If there is even a slight error in your angles, the result may be well off.
 


Thanks guys! That's the formula I wanted! :) thank you! :D Now I can get to the stronghold :DDD
 
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