Find missing coordinate of third vertex in a triangle when one coordinate, two vertices and the angle in the incomplete vertex are given

Question

In triangle $△V_1V_2V_3$ vertices $V_1$, $V_2$, angle $∠ϴ$ and coordinate $y_3$ are known. How do you find $x_3$?

I have revisited triangle theories and similar posts but I did not managed to get to a formula. I feel this is enough information to solve it but not sure how.

Answer 1

On the one hand, the area of the triangle is given by the shoelace formula as

$A = \frac{1}{2} (x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3 ) \hspace{36pt} (1)$

The only unknown on the right hand side is $x_3$.

On the other hand, the square of the area is given by

$A^2 = \dfrac{1}{4} | V_3 V_1 |^2 | V_3 V_2 |^2 \sin^2 \theta \hspace{36pt} (2)$

Explicitly, this is,

$A^2 = \dfrac{1}{4} \sin^2 \theta ( (x_1 - x_3)^2 + (y_1 - y_3)^2 ) ( (x_3 - x_2)^2 + (y_3 - y_2)^2 ) $

Squaring equation $(1)$ and equating the right hand sides results in a quartic (degree 4) equation in $x_3$.

Answer 2

Construct an isosceles triangle with base $\overline{V_1V_2}$ and altitude ${1\over2}\overline{V_1V_2}\cot\theta$. The vertex of this triangle is the center of the circle passing through $V_1V_2V_3$.

As you can see in figure below, two solutions are possible, in general.