geometrytriangles
Let's suppose I have a rectangle divided into three triangles in the following way. No lengths of either the rectangle or triangles are known, only one angle is known.
I would like to know how to calculate all the other angles. Here is a sample image of the problem:
This looks so simple but I have no idea even where to start. Where is the trick?
Hint:
Sides $c$ and $a$ are given, and angle $A$. Notice that there are two possible triangles with side length $a$, and angle $C'$ is the supplement of $C$. Hopes this helps.
Let the two known (shared) vertices be $A = (x_A , 0)$ and $B = (x_B , 0)$, and the two unknown vertices be $C = (x_C , y_C)$ and $D = (x_D , y_D)$.
Let the known variables be $x_A$ and $x_B$, and the distances $$\begin{aligned} L_{AB} &= x_B - x_A \gt 0 \\ L_{AC} &= \left\lVert\overline{A C}\right\rVert \gt 0 \\ L_{BC} &= \left\lVert\overline{B C}\right\rVert \gt 0 \\ L_{AD} &= \left\lVert\overline{A D}\right\rVert \gt 0 \\ L_{BD} &= \left\lVert\overline{B D}\right\rVert \gt 0 \\ \end{aligned}$$
Then, the system of equations that determines the location of $C$ and $D$ is $$\left\lbrace \begin{aligned} L_{AC}^2 &= (x_C - x_A)^2 + y_C^2 \\ L_{BC}^2 &= (x_C - x_B)^2 + y_C^2 \\ L_{AD}^2 &= (x_D - x_A)^2 + y_D^2 \\ L_{BD}^2 &= (x_D - x_B)^2 + y_D^2 \\ \end{aligned} \right .$$ This has four equations and four unknowns, and can be treated as two completely separate systems of equations, each with two unknowns ($x_C$ and $y_C$, and $x_D$ and $y_D$, respectively). The solution is $$\left\lbrace \begin{aligned} \displaystyle x_C &= \frac{ L_{AC}^2 - L_{BC}^2 + x_B^2 - x_A^2 }{2 ( x_B - x_A )} \\ \displaystyle y_C &= \pm \sqrt{ L_{AC}^2 - (x_C - x_A)^2 } = \pm \sqrt{ L_{BC}^2 - (x_C - x_B)^2 } \\ \displaystyle x_D &= \frac{ L_{AD}^2 - L_{BC}^2 + x_B^2 - x_A^2 }{2 ( x_B - x_A )} \\ \displaystyle y_D &= \pm \sqrt{ L_{AD}^2 - (x_D - x_A)^2 } = \pm \sqrt{ L_{BD}^2 - (x_D - x_B)^2 } \\ \end{aligned} \right.$$ If the two triangles extend to the same side, then we can choose the positive signs above (since $y_C \gt 0$ and $y_D \gt 0$). Both right sides for the two $y$ coordinates yield the same answer.
After solving $(x_C , y_C)$ and $(x_D , y_D)$ as above, their distance is obviously $$L_{CD} = \sqrt{ (x_D - x_C)^2 + (y_D - y_C)^2 }$$ Because the distance is necessarily nonnegative, you do not actually need the distance $L_{CD}$ itself; you can just compare the distance squared, $L_{CD}^2$, to the radius squared, $r_C^2$, because nonnegative values compare the same way (less, equal, greater) as their squares do. Expanding the above after squaring yields $$\begin{aligned} L_{CD}^2 &= \left( \frac{ L_{AC}^2 + L_{BD}^2 - L_{AD}^2 - L_{BC}^2 }{ 2 (x_B - x_A) } \right)^2 + \left( \sqrt{ L_{AC}^2 - (x_C - x_A)^2 } - \sqrt{ L_{AD}^2 - (x_D - x_A)^2 } \right)^2 \\ ~ &= \left( \frac{ L_{AC}^2 + L_{BD}^2 - L_{AD}^2 - L_{BC}^2 }{ 2 (x_B - x_A) } \right)^2 + \left( \sqrt{ L_{BC}^2 - (x_C - x_B)^2 } - \sqrt{ L_{BD}^2 - (x_D - x_B)^2 } \right)^2 \\ \end{aligned}$$ Both yield the same solution, to within numerical precision used.
Note that if you had just placed $A = (0, 0)$, ie. $x_A = 0$, you'd have gotten somewhat simpler solutions (and the math would have been easier, too).
Best Answer
Without further information, you cannot determine any of the other angles, other than the right angles (since you are given a rectangle). If we know one other non-right angle, we can compute all the other angles by simply knowing that the angles of a triangle sum to $180^\circ$.