How to find the angle formed by the intersection of three lines forming a triangle?

Question

The problem is as follows:

Find the angle $x$ as indicated in the figure from below:

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&15^{\circ}\\ 2.&20^{\circ}\\ 3.&40^{\circ}\\ 4.&30^{\circ}\\ 5.&45^{\circ}\\ \end{array}$

How exactly the requested angle can be found?. Initially I thought it could had been $60^{\circ}$ but on a closer inspection the figure shows it is not formed by the intersection of those lines.

I remember there's an identity which says that the sum of the exterior angles of a triangle add up to $360^{\circ}$.

In this case it would mean that:

$3\alpha+120+180-\alpha=360$

Then $2\alpha=60$

Then $\alpha=30^{\circ}$

But that's how far I went, can someone help me with a better method?. It would be much helpful to use some diagram or drawing as I can find easier how to find that angle. The approach which I intend to look for is relying on euclidean geometry identities.

Answer 1

$$3\alpha + \angle BAD = 180∘$$

By sum of interior angles of a $\triangle$, $$\angle ABD + 2\alpha + \angle BAD = 180∘$$

Substituting, $$\angle ABD + 2\alpha + {180∘} - 3\alpha = {180∘}$$

$$\angle ABD - \alpha = 0$$ $$\angle ABD = \alpha$$

Therefore, by congruence of opposite angles, $$\alpha + x = {60∘}$$

In $\triangle BDC$, by exterior angle equals sum of interior opposite angles, $$2\alpha = x + \alpha$$

$$x = \alpha$$

Therefore,

$$2\alpha = {60∘} \implies \alpha = {30∘}$$

Therefore, $$x = {30∘}$$

Answer 2

Note also in $\triangle DCB$ you have $2\alpha=\alpha+x$ (the exterior angle is equal to the sum of the two non-adjacent interior angles), so $x=\alpha=30^\circ$.

It really doesn't help that the sketch is quite far off the mark, e.g. the angle $\alpha$ on the image is way bigger than $30^\circ$, $\angle CDB=120^\circ$ should be obtuse but is drawn as acute, etc.