# Mathematics based on triangles

How to find the third cordinate of a triangle , where as other two points are known. and a angle is known.

Lets say , the two points are (0,0) , (600,0) and we need to find the third cordinate . Given an angle 30 degrees.

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If the angle which is known is opposite to the side joining the two known points then the third point lies on an arc of one of two circles whose radius can be determined using the extended sine rule. $$\frac {a}{\sin A}=2R$$

If the known angle is at one end of the known side, then the third point will lie on one of two lines passing through that point and making the requisite angle with the known side.

The number of circles/lines reduces to one when a right angle is involved.

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You have not said which angle is $30$ degrees and even if you had, there would still be several possibilities for the third vertex.

You could consider the possibilities suggested in red in this sketch

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To construct a triangle, we need three pieces of information of these form:

0. a, b, c
1. a, b, C
2. a, B, C


In other cases, there will bu multiple possibilities and only the equation of the points can be derived (locus of points)

In this case, only one side and and angle are known. If we assume that the side, a and opposite angle A are known then the third point would lie on the circles as mentioned by Henry in his answer

If you assume that side a and angle B (or C) are known, then the locus of third point is a set of 4 lines:

   L1: (y-0)/(x-0) = tan(30)
L2: (y-0)/(x-0) = -tan(30)
L3: (y-0)/(x-600) = tan(30)
L4: (y-0)/(x-600) = -tan(30)

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Unfortunately, this is not enough information to find the third point. Imagine the triangle is a right triangle with angles $90^\circ$, $60^\circ$, and $30^\circ$. With a point at $(0, 0)$ and another at $(600, 0)$, there are at least $4$ points alone that are easy to find. For instance, if the third point was at $(0, 346.4)$ we would satisfy your conditions - but it would also satisfy the conditions to put it at $(600, 346.4)$, $(0, -346.4)$, and $(600, -346.4)$. And that's when I specified the other angles, and chose the $30^\circ$ angle to be along the $x$-axis. There are many more triangles that fulfill your requirements.

By the way, I got that $346.4$ number by drawing a right triangle, labeling the adjacent side $600$ and the angle $30^\circ$, and calculating

$\tan(\theta) = O/A$
$\tan(30^\circ) = O/600$
$600\tan(30^\circ) = O$
$346.4 = O$

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