Find the value of angle x (Langley's Adventitious Angles variant) This problem had came to a Facebook post. Given that the left angles are given (with only 40° angle on the upper right triangle,find the angle x . What I did us to make equilateral triangles but it seems I still don't get two angles to get the vale of x . Can anyone give me a solution ???

 A: We can easily solve this by making some clever constructions of some points $X$, and $Y$ in the labelled diagram below;

Note that in the diagram, the triangle you gave is $\triangle ABC,$ and $X$ is produced by reflecting $D$ over $BC$, and $Y$ is constructed so that $BX=BY=BD,$ and $\triangle BXY$ is equilateral.
Note that since $\angle ABX=\angle ABY,$ thus $X$ is the reflection of $Y$ over $AB.$ We know that by Angle Chasing that $\angle BAX=90^\circ-\angle ABC=70^\circ,$ thus by properties of reflection, we know that $\angle BAY=70^\circ$ as well.
But then we have;
$$\angle YAC=\angle YAB+\angle BAX+\angle XAC=70^\circ+70^\circ+40^\circ=180^\circ.$$
Hence, $Y$ lies on $AC.$ We know that $\angle YCB=50^\circ=\angle YBC,$ thus $\triangle YBC$ is isosceles, and thus $YC=YB=YX.$
Hence, $\triangle XYC$ is also isosceles. We will now show that the side length of the lateral sides is equal to the side length of the lateral sides of the isosceles triangle $\triangle XBD,$ note that $XY=XB,$ so the claim is proven.
Now we will prove that $\triangle XYC\cong\triangle XBD$ by showing that $\angle XBD=\angle XYC.$ Note that $\angle XBD=20^\circ$ by the construct, and we have;
$$\angle XYC=90^\circ-\angle BAY=20^\circ.$$
Thus, the claim is proven and the two triangles are congruent by SAS.
We know by Congruence that $XD=XC.$ But since $CD=CX,$ the $\triangle CDX$ is equilateral, and the result follows by noting that;
$$\angle BCD=90^\circ-CDX=30^\circ.$$
A: This is a different problem, but it follows the same principles with this problem. In these problems there are two methods to pick from: Synthetic geometry or trigonometry. Although synthetic solutions are more beautiful, they are harder to come up with. To obtain a synthetic solution you need to create isosceles triangles until you end up with an equilateral triangle that can be used to solve the problem or can be used to obtain a congruency.
A more practical theorem is Trigonometric Ceva's Theorem. (The proof of this theorem is done by applying the law of sines to the three triangles , and multiplying all the obtained expressions.) To apply the theorem just multiply the ratios of the sines of the angles at each vertex respectively:
$$\frac{\sin(40°)}{\sin(70°)}×\frac{\sin(10°)}{\sin(10°)}×\frac{\sin(x)}{\sin(50° - x)}= 1$$
Rearranging:
$$\sin(40)×\sin(x)=\sin(70°)×\sin(50° - x)= 1$$
From here, the answer turns out to be,
$$x=30°.$$
