Given two points and an angle in a triangle, how do I find the other vertices? Assuming the angles are 30 and 60 degrees, and the two vertices are (1, 5) and (5, 1) respectively, how can I get the coordinates of the other vertex?

I'm not good at math and trigonometric functions.
I tried to find a way, but couldn't find it.
Please help me.
 A: You can use the Pythagorean theorem to find the distance between (1,5) and (5,1) then that distance will give you the hypotenuse. From there, you will have your pick between a sine or cosine function to find the other two sides.
A: As you say 30+ 60= 90 so the third angle is 180- 90= 90- this is a right triangle.  The line segment from (1, 5) to (5, 1) is the hypotenuse with length $4\sqrt{2}$.
If we let a be the length of the side opposite the 30 degree angle then $a= 4\sqrt{2} sine(30)= 2\sqrt{3}$.  If we let b be the length of the side opposite the 60 degree angle then $a= 4\sqrt{2}= 2\sqrt{6}$.
The third vertex must lie on the circle with center at (1, 5) and radius $2\sqrt{6}$ so satisfies $(x- 1)^2+ (y- 5)^2= 24$ AND on the circle with center (5, 1) and radius  $2\sqrt{3}$ so satisfies $(x- 5)^2+ (y- 1)^2= 12.
So the vertex, (x, y), satisfies both of those equations.  As ajotate said, there are two points that satisfy both of them.
A: It's not just trigonometry, it's a nice exercise in what used to be called analysis-synthesis: first draw and analyze the situation, then synthesize the fruit of your observations.
Let $A=(1,5)$ and $B=(5,1)$. Then the midpoint of $[AB]$ is $C=(3,3)$. Let $M$ be a point answering what is requested in the statement.
Analysis:there are many ways to prove that $ABM$ is equilateral but it depends on your knowledge.
Synthesis :$M$ necessarily belongs to the intersection of the circles drawn in the figure and reciprocally the two points $C$ and $D$ agree.
Let $(x,y)$ be such a point. So $$\begin{cases} (x-5)^2+(y-1)^2=(2\sqrt2)^2=8\\ ((x-3)^2+(y-3)^2=8 \end{cases}\iff$$
$$\begin{cases} y=x-2\\ (x-3)^2+(y-3)^2=8 \end{cases}\iff \begin{cases} y=x-2\\ y^2-4y+1=0\end{cases}\iff \begin{cases} y=2\mp \sqrt3\\x=4\mp \sqrt3  \end{cases}$$
Conclusion : $D=(4-\sqrt3,2-\sqrt3), C=(4+\sqrt3,2+\sqrt3)$

