How do we find point( b ) on a graph at a specific angle from point( a ). Point( a ) being only value known.

Question

The origin is top-left here because I'm creating this image using SVG. The coordinate system starts in the top left and increases as you move down and/or right. I'm trying to find point b if the orange line is 60 degrees. The only information I have is point a.

Note: My end goal here is to construct an equilateral triangle.

Edit: Angle in picture isn't accurate.

Answer 1

HINT:

Let $ a=50$

$$ ( x-a)^2+(y+a)^2=1 $$ $$y= \sqrt 3(x -a)$$

There are 2 equations and two variables, simplify and solve quadratic for $y$ $$ y^2/3+(y+a)^2=1 $$

Answer 2

We can call define a vector at point $a$ that comes from the origin. Since it is at coordinate location $(50,0)$, it makes sense for the vector to be exactly $(50,0)^T$. Next, applying the rotation matrix (ie multiplying the above vector by the rotation matrix) with an angle of $60$ degrees (ie the angle you have above), we get the new point you desire, point $b$, at $(25,-25\sqrt{3})$.

It's clear this lies on the circle because the norm of the vector with these components equals $50$, which is the radius of your circle.

For more information on rotation matrix, see the following link: https://en.wikipedia.org/wiki/Rotation_matrix

IE, multiplying the vector: \begin{bmatrix}50\\0\end{bmatrix} By the matrix: \begin{bmatrix}\cos(60)&-\sin(60)\\\sin(60)&\cos(60) \end{bmatrix}

Will give you a new vector:

\begin{bmatrix}25\\-25\sqrt{3}\end{bmatrix}

Which holds the coordinates of the point $b$.

Answer 3

Note that

$$ \frac{\vert df\vert}{\vert bf\vert}=\tan30^\circ=\frac{1}{\sqrt{3}}$$

therefore

$$ \vert bf\vert=\sqrt{3}\,\vert df\vert $$

Furthermore,

\begin{eqnarray}\vert bf\vert^2+\vert df\vert^2&=&50^2\\ 4\vert df\vert^2&=&50^2\\ \vert df\vert&=&25\\ \vert bf\vert&=&25\sqrt{3} \end{eqnarray}

So

\begin{eqnarray} b&=&(50+\vert df\vert,50-\vert bf\vert)\\ &=&(75,50-25\sqrt{3} \end{eqnarray}

PS: Note that if you do not want to use $\tan30^\circ$, you can use the fact that $\triangle dbf$ is half of an equilateral triangle to get the proportion between sides $\vert bf\vert$ and $\vert df\vert$.