Finding Coordinates of last Triangle Vertex

[monch]

Hi,
I am trying to find the last vertex coordinates of a triangle given that

Vertex 1 = (2,10)
Vertex 2 = (3,6)

Angle at Vertex 1 = 75.9638 degrees
Angle at Vertex 2 = 70.3462 degrees.


I have tried using the equations based on the length of each side, as well as using the cos dot product rule to try to find x,y but I cannot seem to .

I always end up with the same equations using both methods, hence only giving me 1 equation for 2 unknowns.

 

[HallsofIvy]

You can use the sine law. The distance between vertices 1 and 2 is $\sqrt{(2- 3)^2+ (10- 6)^2}$$= \sqrt{1+ 16}= \sqrt{17}$.
Further, the angles must add to 180 degrees so the angle at vertex 3 is $180- 75.9638- 70.3462= 33.69$ degrees.
So the length, r1, of the side from vertex 1 to vertex 3, the side opposite vertex 2, is given by $r1/sin(70.3462)= \sqrt{17}/sin(33.69)$. Write the equation of the circle with center at vertex 1 with that radius.
And the length, r2, of the side from vertex 2 to vertex 3 the side opposite vertex 1, is given by $r2/sin(75.9638)= \sqrt{17}/sin(33.69)$. Write the equation of the circle with center at vertex 2 with that radius.

Find the points where those two circles intersect by solving the equations simultaneously. There are two solutions, one on either side of the line from vertex 1 to vertex 2.

 

[monch]

Thanks for the reply HallsofIvy,

I end up with only 2 complex roots for some reason when i use these 2 equations

(x-3)^2 + (y - 6)^2 = 52 =(r2^2)
(x-2)^2 + (y-10)^2 = 49 = (r1^2)

i got 10x^2 - 284x +2652 = 0 after substitution.

the answer i am suppose to get for the last coordinate is (9,10)

 

[monch]

oops my mistske, i did some calculation errors.

hmm is there anyway to pick the right coordinates when i end up with 2 solutions?

Both will have the same angle between the vectors , so there's no way to tell which is the coordinate I am looking for?