How to find a point in a Right Triangle given 2 known points, all sides, all interior angles

Question

This triangle is not parallel or vertical, it's in a 2d plane. Distance formula gave me very troubling results, looking to use SOH CAH TOA, particularly a simple method and not a complex method of solving.

Answer 1

Define $$A=\langle11444,1701\rangle$$ $$B=\langle11406,1576\rangle$$ $$\implies A-B=\langle38,125\rangle$$Rotate this $90^\circ$ to get $\langle125,-38\rangle$. Add this to $B$, adjusting for length, to get $C$. $$C = \frac{25}{\sqrt{17069}}\langle125,-38\rangle + B$$ And simplify.

Some explanation: $A-B$ will give you the point $A$ would be at if $B$ was the origin. Drawing a few points on the coordinate plane and rotating them will show you why $\langle x,y\rangle \to \langle y,-x\rangle$ rotates a point $90^\circ$ clockwise. $\langle 125,-38\rangle$ has magnitude ~$131$, which when divided by ~$131$ gives $1$, and which multiplied by $25$ gives $25$. Adding to $B$ will shift the new point right on top of $C$.

Above, $K$ is the rotated version of $A-B$.