A quadrialteral $ABCD$ has $AB = 10$, $\angle A = 50^\circ, \angle B = 120^\circ$, $ BC = x , CD = x + 2 , AD = x + 4 $. Find $x$.
My attempt:
Applying the law of cosines to $\triangle DAC$ and $\triangle ABC$, to obtain
$ AC^2 = (x + 4)^2 + (x+2)^2 – 2 (x + 2) (x + 4) \cos(190^\circ – C) = 10^2 + x^2 – 20 x \cos 120^\circ $
And by applying the law of cosines to $\triangle DAB $ and $\triangle BCD $, I obtained
$ BD^2 = 10^2 + (x + 4)^2 – 20 (x + 4) \cos 50^\circ = x^2 + (x+2)^2 – 2 x (x + 2) \cos C $
It remains to solve the two equations for $x$ and $\cos C$.
Your help on this or through an alternative solution method is much appreciated.
Best Answer
The following is basically your figure, but with the vertical lines $DEF$ and $CG$, plus the horizontal lines $EC$ and $BG$, along with the associated additional labeled points, added to it.
Since $\measuredangle CBG = 60^{\circ}$, then $\measuredangle BCG = 30^{\circ}$, so $\triangle CBG$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle. Thus, we have
$$\lvert BG\rvert = \frac{x}{2}, \;\; \lvert CG\rvert = \frac{\sqrt{3}x}{2}$$
With $\measuredangle ADF = 40^{\circ}$, then
$$\lvert AF\rvert = (x + 4)\sin(40^{\circ}), \;\; \lvert DF\rvert = (x + 4)\cos(40^{\circ})$$
As $CEFG$ is a rectangle, we also get
$$\lvert EC\rvert = \left(10 + \frac{x}{2}\right) - (x + 4)\sin(40^{\circ}), \;\; \lvert DE\rvert = (x + 4)\cos(40^{\circ}) - \frac{\sqrt{3}x}{2}$$
Using these lengths and the Pythagorean theorem on $\triangle DEC$ gives
$$\begin{equation}\begin{aligned} & \left(\left(10 + \frac{x}{2}\right) - (x + 4)\sin(40^{\circ})\right)^2 + \left((x + 4)\cos(40^{\circ}) - \frac{\sqrt{3}x}{2}\right)^2 = (x + 2)^2 \\ & \left(10 + \frac{x}{2}\right)^2 - (20 + x)(x + 4)\sin(40^{\circ}) + (x + 4)^2\sin^2(40^{\circ}) + (x + 4)^2\cos^2(40^{\circ}) - \sqrt{3}x(x + 4)\cos(40^{\circ}) + \frac{3x^2}{4} = x^2 + 4x + 4 \\ & \left(\frac{x^2}{4} + 10x + 100\right) - (x + 4)(x + 20)\sin(40^{\circ}) + (x^2 + 8x + 16) - \sqrt{3}x(x + 4)\cos(40^{\circ}) + \frac{3x^2}{4} = x^2 + 4x + 4 \\ & 2x^2 + 18x + 116 - (x + 4)(x + 20)\sin(40^{\circ}) - \sqrt{3}x(x + 4)\cos(40^{\circ}) = x^2 + 4x + 4 \\ & x^2 + 14x + 112 = (x + 4)((x + 20)\sin(40^{\circ}) + \sqrt{3}x\cos(40^{\circ})) \end{aligned}\end{equation}$$
The Triple-angle formulae section of Wikipedia's "List of trigonometric identities" article indicates $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. Using $\theta = 40^{\circ}$, with $y = \cos(40^{\circ})$ for simpler algebra, along with $\cos(120^{\circ}) = -\frac{1}{2}$, we get $-\frac{1}{2} = 4y^3 - 3y \;\to\; 8y^3 - 6y + 1 = 0$. As indicated in Wolfram Alpha's result by choosing the "Exact forms" button, this is expressed in terms of complex radicals. Thus, I'll use approximate values instead so, continuing with the calculations in the equation above, we get
$$\begin{equation}\begin{aligned} x^2 + 14x + 112 & \approx (x + 4)((x + 20)\times 0.6428 + 1.7321x\times 0.7660) \\ x^2 + 14x + 112 & \approx (x + 4)(1.9696x + 12.8560) \\ x^2 + 14x + 112 & \approx 1.9696x^2 + 20.7344x + 51.424 \\ 0 & \approx 0.9696x^2 + 6.7344x - 60.5760 \end{aligned}\end{equation}$$
Using the quadratic equation, and that $x \gt 0$, we have
$$x \approx \frac{-6.7344 \pm \sqrt{6.7344^2 + 4\times 0.9696\times 60.5760}}{2\times 0.9696} \;\;\to\;\; x \approx 5.161$$
Note this basically matches the $5.16071$ result in David G. Stork's answer obtained using Mathematica.