Given a triangle such as this:
Where $C$, $A$ and $B$ are cartesian coordinates and $a$, $b$, $c$ are the lengths of the sides.
I know that $$C = (b\cos\theta,\;b\sin\theta)$$
where $\theta$ is the angle at vertex $A$.
However, in my case, I do not know $b$, but I do know $c$ and the gradient at both $A$ and $B$.
How can I find $C$?
Best Answer
The equation $C = (b\cos\theta,\;b\sin\theta)$ makes sense only if the coordinates of $A$ are $(0,0)$, so I suppose $A = (0,0)$.
Then $B = (c,0)$.
From a point and the slope of a line, you can derive an equation for the line. So you can easily find equations for lines $AC$ and $CB$.
From the equations of two lines you can find the coordinates of their intersection.
That should be enough to solve the problem. But if the slope of $AC$ is $\tan \alpha$ and the slope of $CB$ is $-\tan \beta$ (where $\alpha$ and $\beta$ are the positive angles at $A$ and $B$, and the slope of $CB$ is negative indicating it slopes down to the right), then the coordinates of $C=(c_x, c_y)$ can be found from the relations
\begin{align} c_y &= c_x \tan \alpha, \\ c_y &= (c - c_x)\tan \beta, \end{align}
as you can confirm by dropping a perpendicular from $C$ to $AB$ and labeling the lengths of the legs and angles $\alpha$ and $\beta$ of the two right triangles so drawn.