0
$\begingroup$

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)

$\endgroup$
3
  • $\begingroup$ You haven't quite specified everything uniquely. Which pair of sides is the angle $\alpha$ measured between? $\endgroup$ Jan 12 '16 at 21:39
  • $\begingroup$ 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. $\endgroup$
    – kernelman
    Jan 13 '16 at 1:35
  • $\begingroup$ and my bad, the second relation is between $\alpha$ and $\gamma$ not $\alpha$ and $\omega$ $\endgroup$
    – kernelman
    Jan 13 '16 at 1:38
1
$\begingroup$

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 $\angle DBC$

enter image description here

If quadrilateral isn't convex, only can happen two situations:enter image description here

enter image description here

Both of them are equivalent to the first case.

$\endgroup$
1
  • $\begingroup$ 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. $\alpha = f(a,b,c,d)$ $\endgroup$
    – kernelman
    Jan 13 '16 at 1:30
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – kernelman
    Jan 16 '16 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.