Problem involving triangle. Find $x$ in the figure

Question

I need to find $x$ in the triangle above.

I tried to do basic things, like sum of a triangle's internal angles $= 180^\circ$ but I only found $2$ equations for $3$ variables

Any help is appreciated

Answer 1

In triangle $ABC$ you have the following angles $\angle A=120^\circ$ (top corner), $\angle B=20^\circ$ (left corner) and $\angle C=40^\circ$ (right corner). Denote the central point with $D$ and introduce lengths $AD=a, BD=b,CD=c$.

By law of sines applied to triangles $ABD,ACD,BCD$:

$$a\sin20^\circ=b\sin10^\circ\tag{1}$$

$$a\sin100^\circ=c\sin(40^\circ-x)\tag{2}$$

$$b\sin10^\circ=c\sin x\tag{3}$$

From (1) and (2):

$$b=\frac{a\sin20^\circ}{\sin10^\circ}$$

$$c=\frac{a\sin100^\circ}{\sin(40^\circ-x)}$$

Replace that into (3):

$$\frac{a\sin20^\circ}{\sin10^\circ}\sin10^\circ=\frac{a\sin100^\circ}{\sin(40^\circ-x)}\sin x$$

$$\sin20^\circ \sin(40^\circ-x)=\sin100^\circ\sin x$$

$$\sin20^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$

$$2\sin10^\circ \cos10^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$

$$2\sin10^\circ \sin(40^\circ-x)=\sin x$$

Sometimes you have to make things more complicated before your are able to jump over the last hurdle: multiply the right side with $1=2\sin30^\circ$.

$$2\sin10^\circ \sin(40^\circ-x)=2\sin x\sin30^\circ$$

$$\cos(-30^\circ+x)-\cos(50^\circ-x)=\cos(x-30^\circ)-\cos(x+30^\circ)$$

$$\cos(50^\circ-x)=\cos(x+30^\circ)$$

For obviously acute angle $x$

$$50^\circ-x=x+30^\circ$$

$$x=10^\circ$$

No calculator needed.

Answer 2

Using the same labeling as the author of the trig proof:

Cut off $BE$ on $BC$, such that $BE=BA$. Then, using the fact that $BD$ is the bisector of $\angle B$, we can easily show that $\triangle ADE$ is equilateral, so that $EA=ED$. Similarly, we can show that $\triangle AEC$ is isosceles (with base angles of $40^\circ$), so that $EA=EC$. Therefore $ED=EC$, so that, in $\triangle EDC, \angle EDC = \angle ECD = x$. But the external angle $BED (=\angle BAD) = 20^\circ$ , therefore $2x = 20^\circ$ , yielding that $x=10^\circ$.