geometrytrigonometry
So i have the figure below:
and i want to find the angle $a$ as shown in green. I have the coordinates of the blue points where
$x_i,y_i = (4,2)$ and $x_j,y_j = (4.3589, 1)$
According to the answer, $a$ is 13.6441 degrees, but i do not know how to work it out.
I used the formula
but I'm not getting the same answer.
Update 2: Here is the graph I got for $(x_{1},y_{1})=( 78. 965,12. 354)$, for the parametric circle $(x(t),y(t))$ centered at $(100,100)$
$$x=100+90.135\cos \left( 1.3527+\pi -t\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -t\frac{\pi }{180}\right) .$$
together with the 4 points $(x(t),y(t))$ for $t=0,90,180,270$
$$(x_{1},y_{1})=(x(0),y(0)),(x(90),y(90)),(x(180),y(180)),(x(270),y(270)).$$
You might use the following equations in a for loop with $k=0$ to $k=359$, step $1$:
$$x=100+90.135\cos \left( 1.3527+\pi -k\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -k\frac{\pi }{180}\right) .$$
to draw the "orbit" with a 1 degree interval.
Update: corrected coordinates of $(x_{1},y_{1})=(140.5,152)$.
You need to consider the new angle and not only the $1{{}^\circ}$ change. The argument of $\cos$ and $\sin$ is this new angle and not $1{{}^\circ}$.
Let $(x_{c},y_{c})=(160,240)$ be the center of the set of circles and $(x_{1},y_{1})=(140.5,152)$. The radius $r$ is
$$\begin{eqnarray*} r &=&\sqrt{\left( x_{c}-x_{1}\right) ^{2}+\left( y_{c}-y_{1}\right) ^{2}} \\ &=&\sqrt{\left( 160-140.5\right) ^{2}+\left( 240-152\right) ^{2}} \\ &=&90.135 \end{eqnarray*}$$
Call $(x,y)$ the new coordinates of $(x_{1},y_{1})$ rotated by an angle of $-1{{}^\circ}=-\dfrac{\pi }{180}$ around $(x_{c},y_{c})$ with a radius $r$. The new angle is $\theta'=\theta -\frac{\pi }{180}$, $\theta $ being the initial angle. Then
$$\begin{eqnarray*} x &=&x_{c}+r\cos \left( \theta -\frac{\pi }{180}\right), \\ y &=&y_{c}+r\sin \left( \theta -\frac{\pi }{180}\right), \end{eqnarray*}$$ where $\theta $ is the angle $\theta =\arctan \dfrac{y_{1}-y_{c}}{x_{1}-x_{c}}:$
$$\begin{eqnarray*} \theta &=&\arctan \frac{152-240}{140.5-160}=1.3527+\pi \text{ rad.}\\ &=&\frac{1.3527\times 180{{}^\circ}}{\pi }+180{{}^\circ}=257. 5{{}^\circ}\end{eqnarray*}$$ Thus $$\begin{eqnarray*} x &=&160+90.135\cos \left( 1.3527+\pi -\frac{\pi }{180}\right)= 138. 96 \\y &=&240+90.135\sin \left( 1.3527+\pi -\frac{\pi }{180}\right) = 152. 35 \end{eqnarray*}$$
This is the solution (be aware that your figure is misleading):
$$ A = (150 - 100, 50 - 100) = (50, -50) $$ $$ B = (180 - 100, 100 - 100) = (80, 0) $$ $$ \cos\Theta = \dfrac{A \cdot B}{|A||B|} = \dfrac{400}{50\sqrt{2} \times 80} = \dfrac{1}{\sqrt{2}} $$ $$ \Rightarrow \Theta = \dfrac\pi4 $$
Best Answer
Your formula is fine, but you must switch your points, because $(x_1,y_1)$ in that formula refers to the point before rotation, which in your case is $(4.3589, 1)$, while $(x_2,y_2)$ refers to the point after rotation, which in your case is $(4, 2)$.