Find the Angle #2

Easy way:

Extend $AD$ and $BC$ to meet at $E$ . Since $BD$ and $AC$ are angle bisectors, their intersection (call it $I$ ) must be the incenter of $\Delta ABE$ . We also have $\angle AEB = 60^{\circ}$ by angle sum of triangle, so $\angle CEI = 30^{\circ}$ . Furthermore, $\angle AIB = 120^{\circ}$ , so $\angle DIC = 120^{\circ}$ . Thus, we have $DICE$ is cyclic, since the opposite angles sum to $180^{\circ}$ . Therefore, $\angle BDC = \angle IEC = 30^{\circ}$ .

I used trigonometry here I assumed a side length equal to 12 in AB then continuously used the sine law.

Relevant wiki: Sine Rule (Law of Sines)

Let $AC$ and $BD$ meet at $O$ . Then $\angle AOB = \angle COD = 180^\circ - 10^\circ - 50^\circ = 120^\circ$ ; $\angle AOD = \angle BOC = \dfrac {360^\circ - 240^\circ}2 = 60^\circ$ ; $\angle OCB = 70^\circ$ and $\angle ADO = 110^\circ$ .

Let $AB = 1$ . By sine rule: $\dfrac {OB}{AB} = \dfrac {\sin \angle OAB}{\sin \angle AOB}$ $\implies OB = \dfrac {\sin 10^\circ}{\sin 120^\circ} \times AB = \dfrac {\sin 10^\circ}{\sin 60^\circ}$ . By sine rule again: $OC = \dfrac {\sin \angle OBC}{\sin \angle OCB} \times OB = \dfrac {\sin 50^\circ}{\sin 70^\circ} \times \dfrac {\sin 10^\circ}{\sin 60^\circ}$ .

Similarly, $AO = \dfrac {\sin \angle ABO}{\sin \angle AOB} \times AB = \dfrac {\sin 50^\circ}{\sin 60^\circ}$ . And $DO = \dfrac {\sin \angle DAO}{\sin \angle ADO} \times AO = \dfrac {\sin 10^\circ}{\sin 110^\circ} \times \dfrac {\sin 50^\circ}{\sin 60^\circ} = \dfrac {\sin 10^\circ}{\sin 70^\circ} \times \dfrac {\sin 50^\circ}{\sin 60^\circ} = OC$ .

Therefore, $\triangle DOC$ is isosceles with $\angle COD=120^\circ$ , then $\angle BDC = \angle ODC = \angle OCD = \dfrac {180^\circ-120^\circ}2 = \boxed{30}^\circ$ .