trigonometry
I'd like to calculate the bearing angle and the distance between heading point and target point with three coordinates for an arduino navigation.
Coordinate system looks something like this.
$(x_1, y_1)$ = heading point, $(x_3, y_3)$ = target point
Angle range should be $-180^\circ$ to $+180^\circ$ so it is possible to continuously adjust bearing angle to zero to drive towards way-point.
Thanks in advance
Let $X$ be the external point. Let $P$ be the center of the circle. Let $Q$ be a point of tangency. Then triangle $\Delta PQX$ is a right triangle. If we let $\theta$ be the angle that $PX$ makes with the horizontal, then the angle of the tangent point with the horizontal is $\theta + \angle XPQ$ and $\theta - \angle XPQ$ respectively. Consequently, the angle that the tangents make with the vertical will also be $\theta \pm \angle XPQ$.
In the figure above, the green angle would be the angle give by $\theta + \angle XPQ$.
If $X = (x_2, y_2)$ and $P = (x_1, y_1)$ then $\theta = \arctan\left(\frac{y_2 - y_1}{x_2 - x_1}\right)$.
If you want the heading to the tangent point, then you will want the red angle (there is another angle for the lower tangent which I have not labelled, it is given analogously) which will be $90^{\circ} - (\theta - \angle XPQ)$.
If you want the angle the tangent makes with the vertical, you will want the blue angle (or the supplement of the blue angle) which is equal to the green angle, $\theta + \angle XPQ$.
You may have to tweak a bit to make sure that everything works out in all quadrants, but this is the basic idea.
Edit If you want $\angle FXQ$ as defined in your comment, then it is simple. Draw a vertical line intersecting $X$. Then the tangent cuts through the vertical line and the $y$ axis as parallel lines. $\angle FXQ$ is then given by the blue angle.
You know red and green dots locations respectively as the present position and target points: $$ (x_1,y_1),(x_2,y_2)$$ Distance between them: $$ d= \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$$ Use $atan2$ to find direction (automatically sensitive to signs of quadrant you are in, and the target location) $$ \theta_{relative}= atan2 \dfrac{y_2-y_1}{x_2-x_1}$$ then extra pinging angle: $$ \theta_{Orange Heading}(35^{\circ}) + 90^{\circ}-\theta_{relative} $$
Please check the above, because trigonometrical angles are measured counterclockwise and the heading rotation you gave are clockwise.
Best Answer
The three points form a triangle, as in the figure below:
You know the coordinates of the three vertices: $H(x_H,y_H)$, $A(x_A,y_A)$, and $T(x_T,y_T)$. The goal is to compute the angle between $\vec{AT}$ and $\vec{AH}$. I write these as vectors because you want an oriented angle, starting from the segment $\vec{AT}$ and increasing counterclockwise towards $\vec{AH}$.
The derivation below will result in an angle in the range $0 \le \theta < 360^\circ$. You can adjust the range to $-180^\circ \le \theta < 180^\circ$ by simply subtracting $180^\circ$ from the value we'll compute below.
We'll obtain the angle by making use of the dot and cross products of the vectors $\vec{AT}$ and $\vec{AH}$ to find the values of $\cos\theta$ and $\sin\theta$, respectively. The dot product between two vectors in $\mathbb{R}^2$ is a number which can be computed in two ways: $$ \vec{a}\cdot\vec{b} \equiv a_x\,b_x + a_y\,b_y = |\,\vec{a}\,|\,|\,\vec{b}\,|\cos\theta $$
Similarly, the cross product between two vectors in $\mathbb{R}^2$ is also a number which can be computed in two ways: $$ \vec{a}\times\vec{b} \equiv a_x\,b_y - a_y\,b_x = |\,\vec{a}\,|\,|\,\vec{b}\,|\sin\theta $$
Dividing the result from the cross product by the result for the dot product, we can compute $\tan\theta$: $$ \tan\theta \equiv \frac{\sin\theta}{\cos\theta} = \frac{a_x\,b_y - a_y\,b_x}{a_x\,b_x + a_y\,b_y} $$ Note that we can't perform that division when $\cos\theta$ (the denominator) is zero, that is, when $\theta = 90^\circ$ or when $\theta = 270^\circ$. Now for the actual computation.
The case when both $c = 0$ and $s = 0$ cannot happen unless the three points $A$, $T$, and $H$ coincide. As I said at the top, this results in $0 \le \theta < 360^\circ$ but you can adjust the range to $-180^\circ \le \theta < 180^\circ$ by subtracting $180^\circ$ from the angle computed above.