Calculate the third point of an obtuse triangle

Question

I have the x,y coordinates of point A, point B, the distance AB, the distance BC, and the angle at B, which is more than 90 degrees. How do I calculate point C?

It's been years since varsity, so my math is very rusty. Please do not assume much :-)

score 2 · Answer 1 · edited Sep 18 '18 at 15:36

2

First, use cosine theorem to find $AC$. $AC^2=AB^2+BC^2-2AB\cdot BC\cdot \cos B$.
Let $AC=b$, $AB=c$, $BC=a$ and points have coordinates $A(x_A, y_A)$, $B (x_B, y_B), C (x_C, Y_C)$. Then we have this system of equations to find coordinates of point $C$: \begin{cases} (x_C-x_B)^2+(y_C-y_B)^2=a^2 \\ (x_C-x_A)^2+(y_C-y_A)^2=b^2 \end{cases}

We have two unknowns $x_C$ and $y_C$ and two equations.
Basically, what we are doing here is drawing two circles, one has $A$ as center and radius $AC$ and the other one has $B$ as a center and radius $BC$. Point $C$ will be the intersection of these circles. Be aware that you will find two solutions as $C$ has two possible locations based on the information.

Solving the system: \begin{cases} x_C^2-2x_B x_C+x_B^2+y_C^2-2 y_B y_C+y_B^2=a^2 \\ x_C^2-2x_Ax_C+x_A^2+y_C^2-2y_Ay_C+y_A^2=b^2 \end{cases}

Subtracting second equation from the first: \begin{cases} x_C(2x_A-2x_B)+y_C(2y_A-2y_B)=a^2-b^2+x_A^2+y_A^2-x_B^2-y_B^2 \\ x_C^2-2x_Ax_C+x_A^2+y_C^2-2y_Ay_C+y_A^2=b^2 \end{cases}

Expressing $x_C$ in terms of $y_C$:

\begin{cases} x_C=\frac{a^2-b^2+x_A^2+y_A^2-x_B^2-y_B^2-y_C(2y_A-2y_B)}{2x_A-2x_B} \\ x_C^2-2x_Ax_C+x_A^2+y_C^2-2y_Ay_C+y_A^2=b^2 \end{cases}

All that's left is to substitute $x_C$ in the second equation and we'll get a quadratic equation with one unknown $y_C$.

edited Sep 18 '18 at 15:36

Peet Brits

103
5

answered Sep 16 '18 at 02:52

Vasili

10,690

I need a bit more explanation to understand this one... – Peet Brits Sep 16 '18 at 13:42
@PeetBrits: I added additional explanation, let me know if this helps you to solve the problem – Vasili Sep 17 '18 at 00:26
So, first formula is law of cosines (without the last BC part - https://en.wikipedia.org/wiki/Law_of_cosines ), which calculates the side lengths to give me AC. The next formula calculates the distance between two points, but I need xC and yC. I'm not sure how to change this formula to get xC and yC. – Peet Brits Sep 17 '18 at 03:49
1

@PeetBrits: I put some additional steps regarding finding $x_C$ and $y_C$. It really not that complicated when you are working with real numbers. – Vasili Sep 17 '18 at 10:32
Okay, thanks! Just one last thing (sorry, I'm a bit rusty!). When I do the substitution at "Subtracting second equation from the first" I get an additional xB^2 - xA^2 + yB^2 - yA^2. Did I miss something? – Peet Brits Sep 17 '18 at 16:14
@PeetBrits: You are right, it was me who was missing something :) I updated accordingly – Vasili Sep 18 '18 at 13:00

score 1 · Accepted Answer · answered Sep 16 '18 at 02:57

1

You can use vectors to find the position of $C$ relative to the point $A$. So it will suffice if you can calculate the vector $\vec{AC}$.

answered Sep 16 '18 at 02:57

Soroush khoubyarian

1,072

So, this translates to cX = bX + BC * (-cos(theta - alpha)) and cY = bY + BC * sin(theta - alpha) – Peet Brits Sep 17 '18 at 04:45

Calculate the third point of an obtuse triangle

2 Answers2