Solve irregular quadrilateral given 2 angles and 3 sides

Question

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.

Answer 1

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.

Answer 2

Assume $\pi/2<\alpha<2\pi$ as suggested by the drawing. Extend the sides of lengths $AB$ and $FE$ till the intersection at $D$ and draw $EC\perp AD$. $$ BC = c\cos(\pi-\alpha) = -c\cos\alpha \\ CE = c\sin(\pi-\alpha) = -c\sin\alpha $$ From similarity of $\triangle ADF$ and $\triangle CDE$ we get $$ -\frac{a}{c\sin\alpha} = \frac{2a+ED}{ED}\\ ED=-\frac{2ac\sin\alpha}{a+c\sin\alpha} $$ Hence $$ FD = FE + ED = \frac{2a^2}{a+c\sin\alpha} $$ Now $$ \cos\gamma=\frac{AF}{FD}=\frac{a+c\sin\alpha}{2a}\\ \beta = \pi - \gamma - (2\pi - \pi/2 - (\pi-\alpha)) = \frac{\pi}{2} + \alpha - \gamma $$ Finally $$ b = AC - BC = 2a\sin\gamma + c\cos\alpha $$ It does not seem like there is a nice answer in terms of $a$, $c$ and $\alpha$ alone. A similar construction should work in case $0<\alpha<\pi/2$.