What's 7 Doing In There?

$x=\dfrac{150}{7}$

Note: Please check Albert Lau's proof using a 42-gon, based on concurrency of 3 diagonals

Albert Lau's 42-gon proof

Albert Lau's approach in using n-gon is more general and can be used for other such adventitous quadrilaterals. See his other posted adventitous quadrilateral problems.

Adventitous Quadrilateral 2

Adventitous Quadrilateral 3

Adventitous Quadrilateral 1

Here, another geometric proof will be given. For an arbitrary angle $x$ , construct irregular pentagon $ABCDE$ as per the following figure, WITHOUT drawing line $AD$ with the angles $30-x$ inside triangle $\Delta AED$ . This can be done in a straightforward way.

It can be readily seen that when line $AD$ is drawn, triangle $\Delta AED$ is isoceles, and that angles $\angle EAD = \angle EDA = 30-x$ . From this, angle $\angle BCA = 60-\angle CAD$

Now, alternatively by trigonometric methods, first, we start with a true equation, and work from there, making use of sum/product of Sines and Cosines

$2\cdot \frac { 1 }{ 2 } -1=0$

$2Cos \left (\frac { 420 }{ 7 } \right)-1=0$

$2Cos\left( \frac { 420 }{ 7 } \right) Sin\left( \frac { 480 }{ 7 } \right) -Sin\left( \frac { 480 }{ 7 } \right)=0$

$Sin\left( \frac { 900 }{ 7 } \right) +Sin\left( \frac { 60 }{ 7 } \right) -Sin\left( \frac { 480 }{ 7 } \right) =0$

$Sin\left( \frac { 900 }{ 7 } \right) -2Cos\left( \frac { 270 }{ 7 } \right) Sin\left( \frac { 210 }{ 7 } \right) =0$

$Sin\left( \frac { 270 }{ 7 } \right) Sin\left( \frac { 900 }{ 7 } \right) -2Sin\left( \frac { 270 }{ 7 } \right) Cos\left( \frac { 270 }{ 7 } \right) Sin\left( \frac { 210 }{ 7 } \right) =0$

$Sin\left( \frac { 270 }{ 7 } \right) Sin\left( \frac { 900 }{ 7 } \right) -Sin\left( \frac { 540 }{ 7 } \right) Sin\left( \frac { 210 }{ 7 } \right) =0$

Now refer to the figure below, with more angles put in as easily determined, so that triangle $\Delta ACD$ is established, and we let $CD=1$ . Then we form another triangle $\Delta ABC'$ such that $\angle ABC'= \angle CAD+90$ , and $BC'=1$ . We don't know if $BC'$ actually meets at vertex $C$ , but we shall find out.

We then determine the length of $AC$ , $AC'$ of each, using Law of Sines.

$\dfrac { 1 }{ Sin\left( \frac{210}{7} \right) } =\dfrac { AC }{ Sin\left( \frac { 900 }{ 7 } \right) }$

$\dfrac { 1 }{ Sin\left( \frac { 270 }{ 7 } \right) } =\dfrac { AC' }{ Sin\left( \frac { 540 }{ 7 } \right) }$

From the equation derived from above, we see that $AC=AC'$ , and from there we can determine that $\angle BCA = \dfrac{150}{7}$ . And once it is seen that $\Delta BCD$ is isoceles, we can draw line $BD$ back again with all angles intact.

See Sum-to-Product Identities
See Product-to-Sum Identities

For any $\theta$ (here in this problem $\theta =\frac{270}{7}$ ), the following are true for some $L$

$\dfrac { 1 }{ Sin( 30 ) }=\dfrac { L }{ Sin( \theta + 90 ) }$

$\dfrac { 1 }{ Sin( \theta) } =\dfrac { L }{ Sin( 2\theta) }$

since if

$2=\dfrac { L }{ Cos( \theta ) }$

then

$\dfrac { 1 }{ Sin( \theta) } =\dfrac { L }{ 2Sin( \theta)Cos(\theta)} = \dfrac { L }{ Sin( 2\theta) }$

For a less trivial example which doesn't make use of this identity, check this out

If angles $a, b, c, d$ at the bottom of the quadrilateral are known, then

$x=ArcCot\left( \dfrac { Sin(a+b+c)Sin(c+d) }{ Sin(a)Sin(c)Sin(b+c+d) } -Cot(a) \right)$

$a=\angle BAC$
$b=\angle CAD$
$c=\angle ADB$
$d=\angle BDC$
$x=\angle BCA$

If $a, b, c, d$ are all rational angles, and $x$ turns out to be rational also, then all angles in the quadrilateral are rational.

Alternatively (per Ceva's Theorem, thanks Dong kwan Yoo),

$Sin(x)Sin(d)Sin(a+b)-Sin(a)Sin(c)Sin(b+c+d-x)=0$

or

$Sin(a+b+d+x)+$
$Sin(a+b-d-x)+$
$Sin(a-b-d+x)+$
$Sin(-a-b+d-x)+$
$Sin(-a+b+2c+d-x)+$
$Sin(-a-b-2c-d+x)=0$

The Question: For any quadrilateral with all rational angles, is it always possible to resolve this equation, through repeated application of Sum-to-Product and Product-to-Sum identities?

I am bad with math.

This solution is inspired by Michael Mendrin's solution on a similar problem.

This solution will be attempted without trigonometry as an added challenge.

for simplicity, let $u=\frac{30^{\circ}}{7}, \ 42u=180^{\circ}$

we start by constructing a regular 42-gon and connect 6 diagonals. any angles on vertices are founded by counting number of subtended sides. the rest can be found by using triangles. by comparing the angles, $x=5u=\frac{150^{\circ}}{7}$ . unfortunately this proof is incomplete without proving concurrency of 3 (red) lines.

we will complete the proof by constructing $AD, ED$ , and try to prove $BCDF$ is collinear. on left diagram, we construct 5 diagonals as shown. connect $CD$ . it can be shown that $ACDE$ is a kite, hence $\angle ADC=7u$ .

on right diagram, we construct 3 diagonals as shown. $\angle ADB=7u$ .

since both $CD$ and $BD$ makes the same angle $7u$ from $AD$ , then $BCDF$ must be collinear, proving concurrency of lines $BF,AD,ED$ , which completes the proof.

Just think of two same base triangle ABC and DBC with common base BC.

in triangle ABC,
angle ABC=30+270/7,
angle ACB=30 and angle BAC=180-(ABC+ACB).

In triangle DBC,
angle DBC=270/7,
angle DCB=30+330/7,
angle BDC=180-(DBC+DCB).

Suppose BC=1 unit, then try to find AC and CD,
You know the angle ACD ,
now it is easy for you to find angle ADC from triangle ADC.
Then find ADC-BDC
You will find ADB which is 150/7

x = $\frac{150}{7}$

We can solve this problem by using slopes. First notice that in order to make a line have a certain angle from the x axis, you make the tangent of the desired angle the slope. For example:

We can model this figure with equations using this technique:

We can then find the slope between A and B, which after a lot of ugly algebra is equal to tan(120/7). The other "half" of angle A is 30 degrees, which can be found from the slope between point A and the origin.

Angle A = 330/7

Adding these two angles yields (\frac{330}{7}) degrees for angle A.

The sum of all interior angles except for X and A can be found through some finagling of angles, and turns out to be $\frac{2040}{7}$ . Since the inside angles of a quadrilateral add up to 360:

2040/7 + A + X = 360

A + X = 480/7

330/7 + X = 480/7

X = 150/7