Finding angles of a quadrilateral

Question

There is a quadrilateral. Length of all $4$ sides are known (lets say $a,b,c,d$).

All $4$ angles are $\leq 180^{\circ}$, but their exact value is unknown. lets say the four angle names are $\alpha,\beta,\gamma,\omega$ (in this order)

refer this fig: https://i.stack.imgur.com/ejefG.jpg (img courtesy: @sinbadh)

$\alpha$ and $\beta$ has following relation (1st):

$\alpha = \beta/2 + 90^{\circ}$

Is this info enough to find a unique solution for $\alpha,\beta,\gamma,\omega$

If yes, could you help provide their solution.

If no, lets make a relation (2nd) between $\alpha$ and $\gamma$:

$\gamma = 180^{\circ} - \alpha$

Now is it possible to find a unique solution for $\alpha,\beta,\gamma,\omega$

In short i need $\alpha = f(a,b,c,d)$

----- UPDATE -----

With the help of @sinbadh's answer posted below, on using law of cosine I was able to find $\alpha$ and $\gamma$ as $f(a,b,c,d)$ only if the second relation ($\gamma = 180 - \alpha$) is true.

However, it would help me more if I can find the angles with only 1st relation (not 2nd relation)

Answer 1

In figure, by Cosine's Law we know $AB$. Then, by the same law, we know $\angle ADB$.Finally, as it is a convex quadrilateral (cuase all angles are not mayor than 180), sum of all angles are 360. Those, we get $\angle DBC$

If quadrilateral isn't convex, only can happen two situations:

Both of them are equivalent to the first case.

Answer 2

The possible values of $\cos \alpha$ are given by the real roots on the interval $[-1,1]$ of the cubic polynomial $$ P(x) \equiv 8 b d x^3 - 4 ab x^2 + (2 ad - 6bd)x - (a^2 + b^2 - 2 ab - c^2 + d^2) = 0, $$ if any such roots exist.

To show this, set up a Cartesian coordinate system such that the angle $\alpha$ is at the origin and the side $a$ lies along the $x$-axis. We can derive the coordinates of the vertex opposite $\alpha$ in two ways. By considering displacements along sides $a$ and $b$, the coordinates of the opposite point are $$ (a - b \cos \beta, b \sin \beta) = (a + b \cos (2 \alpha), - b \sin (2 \alpha)), $$ where we have used the relationship $\beta = 2 \alpha - \pi$ in the second step. On the other hand, if we consider the displacements along sides $c$ and $d$, the coordinates of this same point must be $$ \left(d \cos \alpha + c \cos(\omega + \alpha - \pi), d \sin \alpha + c \sin(\omega + \alpha - \pi)\right) = \left(d \cos \alpha - c \cos(\psi), d \sin \alpha - c \sin(\psi)\right), $$ where $\psi \equiv \alpha + \omega$. These two coordinates must be equal to each other, meaning that we have two equations and two unknowns ($\alpha$ and $\psi$): \begin{align*} a + b \cos (2 \alpha) &= d \cos \alpha - c \cos \psi \\ - b \sin (2 \alpha) &= d \sin \alpha - c \sin \psi \end{align*} Isolating $\psi$ on one side of each equation, squaring each equation, and adding them together then yields $$ (a + b \cos (2 \alpha) - d \cos \alpha)^2 + (b \sin (2 \alpha) + d \sin \alpha)^2 = c^2, $$ which, when expanded out, yields $$ a^2 + b^2 + d^2 + 2ab \cos (2\alpha) - 2ad \cos \alpha + 2bd [-\cos (2 \alpha) \cos \alpha + \sin (2 \alpha) \sin \alpha] = c^2. $$

Now let $x = \cos \alpha$. This means, in particular, that $\cos (2\alpha) = 2x^2 - 1$ and \begin{align*} -\cos (2 \alpha) \cos \alpha + \sin (2 \alpha) \sin \alpha &= -(2 \cos^2 \alpha - 1) \cos \alpha + 2 \sin^2 \alpha \cos \alpha \\ &= -(2x^2 - 1)x + 2(1 - x^2)x \\ &= - 4x^3 + 3x. \end{align*} Thus, plugging this in, we get that $x = \cos \alpha$ must satisfy the polynomial $$ a^2 + b^2 + d^2 + 2ab (2x^2 - 1) - 2ad x + 2bd (- 4x^3 + 3x) = c^2, $$ from which the condition $P(x) = 0$ (with $P(x)$ defined as above) follows.

This is about as far as it's worth going analytically, to be honest. Closed-form solutions to cubic polynomials exist, but they're notoriously hard to work with analytically. However, it's not too hard to derive a necessary condition that there be at least one root in the interval $[-1,1]$. We have $$ P(-1) = c^2 - (a + b + d)^2 $$ and $$ P(1) = c^2 - (a + b - d)^2 $$ But for the quadrilateral to exist in the first place, we must have $ c < a + b + d$; thus, $P(-1) < 0$. If $P(1) \geq 0$, there will necessarily be at least one root of $P$ in the interval $[-1,1]$; this occurs when $$ c \leq | a + b - d|. $$ However, this is only a sufficient condition for a solution to exist; I haven't been able to prove that it is a necessary condition as well.