Given triangle ABC, how to move point B to a certain angle given that its new location lies within the direction of its old altitude. I have a 2D coordinate system for 3 known points $A$, $B$, $C$. Given that I can only move point $B$, how can I compute for its new coordinate with a certain angle $\theta$ considering that its new location can be found along the axis of its old altitude?
Fig. 1: Point B having 110° angle

Fig. 2: New location for B where $\theta = 90°$

In the example above, Fig. 1 shows that the initial angle at point $B$ is $110°$. If input $\theta = 90°$, then point $B'$ will be located somewhere along the x-axis (where the altitude in the current example is located).
Input $\theta$ can be any angle from $1°$ to $180°$. Is there a way to compute for the coordinate of B' given that we know the direction of the altitude of B?
 A: Building on @AxelKemper's suggestion, you can solve for the position of $\mathbf{B'}$ using basic trig functions.
Illustrated example of triangle solution
See above image for details.
Suppose you have a simplified case where the triangle $\Delta\mathbf{ABC}$ rests along the X-axis, and point $\mathbf{A}$ is located at the origin. You are given the measures of the angles $\mathbf{{\lt}BAC}$, $\mathbf{{\lt}ABC}$, $\mathbf{{\lt}ACB}$, and $\mathbf{{\lt}AB'C}$, as well as the lengths of line segments $\mathbf{\overline{AB}}$, $\mathbf{\overline{BC}}$, and $\mathbf{\overline{CA}}$. What is the Y-coordinate of $\mathbf{B'}$?
The point $\mathbf{V}$ indicates the vertical height of $\mathbf{B}$ and $\mathbf{B'}$ with respect to the line $\mathbf{\overline{AC}}$, i.e. $\mathbf{{\lt}AVB}$ and $\mathbf{{\lt}AVB'}$ are right angles.
The point $\mathbf{M}$ is the centers of the circle circumscribed around triangle $\Delta\mathbf{AB'C}$.
The point $\mathbf{D}$ is the midpoint of $\mathbf{\overline{AC}}$, so line $\mathbf{\overline{DM}}$ bisects line $\mathbf{\overline{AC}}$ and angle $\mathbf{{\lt}AMC}$.
$\mathbf{{\lt}AMC} = 2 * \mathbf{{\lt}AB'C}$
Using the inscribed angle theorem, we can determine that the measure of angle $\mathbf{{\lt}AMC}$ is twice that of $\mathbf{{\lt}AB'C}$.
$\mathbf{\overline{AM}} = 
\mathbf{\overline{CM}} = 
\mathbf{\overline{B'M}}$
Since $\mathbf{A}$, $\mathbf{B'}$, and $\mathbf{C}$ are all points along circle $\mathbf{M}$, we know that these three radii have equal length.
$\mathbf{\overline{AM}} =  \frac{\mathbf{\overline{AC}}}{2 sin \mathbf{{\lt}AB'C}}$
The line $\mathbf{\overline{DM}}$ bisecting $\mathbf{\overline{AC}}$ will pass through and also bisect $\mathbf{{\lt}AMC}$ since $\mathbf{{\Delta}AMC}$ is isosceles, yielding a right triangle $\mathbf{{\Delta}ADM}$ and allowing us to use trigonometric functions.
$\mathbf{\overline{AV}} = \mathbf{\overline{AB}} cos \mathbf{{\lt}BAV}$
The distance from $\mathbf{A}$ to the vertical line $\mathbf{\overline{VB}}$ passing through $\mathbf{B}$ and $\mathbf{B'}$.
$\mathbf{\overline{AD}} = \frac{\mathbf{\overline{AC}}}{2}$
Since $\mathbf{\overline{MD}}$ bisects $\mathbf{\overline{AC}}$.
$\mathbf{\overline{VH}} = 
\mathbf{\overline{MD}} = \sqrt{ 
\mathbf{\overline{AM}}^2 -
\mathbf{\overline{AD}}^2}$
The vertical height of triangle $\mathbf{{\Delta}AMC}$, which can be calculated using the Pythagorean Theorem.
$
\mathbf{\overline{HM}} =
\mathbf{\overline{VD}} =
\mathbf{\overline{AD}} -
\mathbf{\overline{AV}}
$
Let $\mathbf{H}$ be the point along $\mathbf{\overline{VB'}}$ where a line perpendicular to $\mathbf{\overline{VB'}}$ passes through $\mathbf{M}$ (black line on diagram). This line, $\mathbf{\overline{HM}}$, is the vertical height of triangle $\mathbf{{\Delta}VMB'}$, and is equal in length to $\mathbf{\overline{VD}}$.
$
\mathbf{\overline{HB'}}=
\sqrt{
\mathbf{\overline{MB'}}^2-
\mathbf{\overline{HM}}^2}
$
The right triangle $\mathbf{{\Delta}B'HM}$ allows us to use the Pythagorean theorem to determine the length of $\mathbf{\overline{HB'}}$ using $\mathbf{\overline{HM}}$ and the radius of the circumscribed circle, $\mathbf{\overline{MB'}}$.
$
\mathbf{\overline{VB'}}=
\mathbf{\overline{VH}}+
\mathbf{\overline{HB'}}
$
Finally, the height of the right triangle $\mathbf{{\Delta}AVB'}$ can be solved by adding the lengths of the two other legs, QED.
I believe this is what you were trying to solve in the image above, so from here it would be fairly straightforward to calculate the coordinates of and distances between each of the points $\mathbf{A}$, $\mathbf{B'}$, and $\mathbf{C}$.
