I have some eight points with defined coordinates in 3D space. These are from a straigth but bent and connected with matlab code as in Figure 1. 
My challenge is that I need to compute the angle that exists between each consecutive 3D points in the bent line after folding the straight line to reach a defined target point. I have added Figure 1 to show the angles I am referring to. How can I calculate those angles by trigonometry?
[Math] Angle of Elevation (Depression) Between two 3D points
analytic geometrygeometrykinematicsroboticstriangles
Related Solutions
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.
Draw circle $c$ of center $C$ and radius $CA$, meeting side $CB$ at $D$ (I'm assuming $AC<BC$).
Draw circle $b$ of center $B$ and radius $BD$, meeting side $AB$ at $E$.
Draw line $ED$: its second intersection with circle $c$ is point $X$ such that $CX=CA$ (obvious) and $CX\parallel AB$ (because triangles $CDX$ and $BDE$ have the same angles).
Draw line $AX$, meeting side $BC$ at $X'$, which is a vertex of the rhombus.
Draw circle $d$ of center $X'$ and radius $X'D$, meeting line $ED$ at $F$; line $X'F$ is parallel to $AB$, because triangles $X'DF$ and $BED$ have the same angles.
Draw line $AD$, meeting circle $d$ at $G$; line $X'G$ is parallel to $AC$, because triangles $X'DG$ and $CDA$ have the same angles.
- Draw lines $X'F$ and $X'G$ to complete the rhombus.
Best Answer
I am indebted in this solution to @Nominal Animal that has made me the fundamental remark that using cross product couldn't do in all situations.
Let us recall the dot product formula:
$$\tag{1}\vec{W} \cdot \vec{W'}=\|\vec{W}\|\|\vec{W'}\|\cos(\theta) \ \ \ \iff \ \ \ \cos(\theta) = \dfrac{\vec{W} \cdot \vec{W'}}{\|\vec{W}\|\|\vec{W'}\|}$$
where $\theta$ is the angle between vectors $\vec{W}$ and $\vec{W'}$.
Here is the way one can proceed for finding the angles:
Compute the consecutive differences between your $n=8$ consecutive points, in other terms, compute the components of the $n-1=7$ vectors $U_k:=\overrightarrow{P_kP_{k+1}}=P_{k+1}-P_k$ ($k=1,2,\cdots n-1$, differences component by component, of course).
Compute the norms (lengths) of the $U_k$s.
Compute the $n-2=6$ dot products $d_k=U_k \cdot (-U_{k+1})$ (minus sign is important). Then, using formula $(1)$:
Compute the $a_k=\frac{d_k}{\|U_k\|.\|U_{k+1}\|}$ (the point meaning "ordinary product" between numbers.)
Last step, the looked for angles are the $\cos^{-1}(a_k)$ of these numbers (refer to relationship $(1)$.)
Recall: the other formula for dot product: $\pmatrix{a\\b\\c} \cdot \pmatrix{d\\e\\f}=ad+be+cf.$