Given a follwing irregular quadrilateral:
We know 3 sides (a,2a,c) and 2 angles (one right angle and alpha).
Need to find side b and two other angles.
[Math] Solve irregular quadrilateral given 2 angles and 3 sides
geometryquadrilateral
Related Solutions
As stated the problem can have many solutions.
For example:
Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.
Figure 1
The angle $b$ depends on $a$ as the expression:
$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$
So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.
Figure 2
There is no such quadrilateral. For draw the hypotenuse of the right triangle with legs $882.9$ and $80$. This hypotenuse has length $l$ greater than $882.9$.
But $293.3+576.8\lt 882.9\lt l$, contradicting the triangle inequality: the sum of any two sides of a triangle is greater than the third side.
Best Answer
From the picture, $\angle {DCF}=180°-α$ and by construction $\angle F=90°$, thus $DF=c*sin(180°-α)$. $ED=EF-DF$, which is $ED=a-c*sin(180°-α)$. Again, by construction $\angle {B}=90°$, so $\angle {DBE}=90°-γ$ and $sin(\angle {DBE})=\frac{a-c*sin(180°-α)}{2a}$, from which we get $\angle {DBE}=sin^{-1}(\frac{a-c*sin(180°-α)}{2a})$. Because $\angle {DBE}=90°-γ$, $γ=90°-\angle {DBE}$ which is $γ=90°-sin^{-1}(\frac{a-c*sin(180°-α)}{2a})$.
Now, $\angle {EDB}=90°-(90°-γ)=γ$ and $\angle {CDF}=90°-180°+α=α-90°$ ⇒ $β=180°-\angle {EDB}-\angle {CDF}=180°-(90°-sin^{-1}(\frac{a-c*sin(180°-α)}{2a}))-(α-90°)=180°+sin^{-1}(\frac{a-c*sin(180°-α)}{2a})-α$.
Let's take care of $b$. $AF=BE$, from $\triangle {BED}$ we get $BE=2a*sinγ$
(because $\angle {EDB}=γ$) and finally, $b=AF-CF$, where $CF=cos(180°-α)*c$, thus $b=2a*sinγ-cos(180°-α)*c$
This solution seems bit complicated, but I couldn't come up with the easier one, hope it helps.