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: http://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)

• You haven't quite specified everything uniquely. Which pair of sides is the angle $\alpha$ measured between? Jan 12 '16 at 21:39
• i apologize for the poor description. but $(a,b,c,d) are just random names of sides. e.g. in the answer posted by sinbad below, the 1st quadrilateral is the one i am interested in. Jan 13 '16 at 1:35 • and my bad, the second relation is between$\alpha$and$\gamma$not$\alpha$and$\omega$Jan 13 '16 at 1:38 2 Answers In figure, by Cosine's Law we know$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$If quadrilateral isn't convex, only can happen two situations: Both of them are equivalent to the first case. • Thanks for the comment sinbadh. yes, the quadrilateral is indeed convex as evident from each angles <=180. I need to figure out values of angles e.g.$\alpha = f(a,b,c,d)$Jan 13 '16 at 1:30 The possible values of$\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$\cos (2\alpha) = 2x^2 - 1and \begin{align*} -\cos (2 \alpha) \cos \alpha + \sin (2 \alpha) \sin \alpha &= -(2 \cos^2 \alpha - 1) \cos \alpha + 2 \sin^2 \alpha \cos \alpha \\ &= -(2x^2 - 1)x + 2(1 - x^2)x \\ &= - 4x^3 + 3x. \end{align*} Thus, plugging this in, we get thatx = \cos \alpha$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.

• I am very much thankful to u for all the effort you made for the explanation. I understood your point very well. I used, wolframalpha to solve for x, and confirmed there is not a unique solution for it. Well, this implies I can't use only relation 1 to find concrete answers for my work & I have to use relation 2 as well. Since Sinbadh's hint helped me to get the ans, I am marking it as "correct answer". Although I want to mark ur answer as "correct answer" as well, but I can't. I don't have enough reputation to upvote your answer either. I apologize sincerely. Jan 16 '16 at 10:40