Chasing an angle

Compute some unknown angles.

$\angle BDC=180-20-40=120$

$\angle ADC=180-10-30=140$

$\angle ADB=360-120-140=100$

Let $AC=1$ . Apply sine law on $\triangle ADC$ .

$\dfrac{AD}{\sin 30}=\dfrac{1}{\sin 140}$ $\implies$ $AD=\dfrac{\sin 30}{\sin 140}$

$\dfrac{CD}{\sin 10}=\dfrac{1}{\sin 140}$ $\implies$ $CD=\dfrac{\sin 10}{\sin 140}$

Apply sine law on $\triangle BDC$ .

$\dfrac{BC}{\sin 120}=\dfrac{CD}{\sin 20}$ $\implies$ $BC=\dfrac{\sin 120}{\sin 20}\left(\dfrac{\sin 10}{\sin 140}\right)$

Apply cosine law on $\triangle ACB$ .

$AB^2=\left(\dfrac{\sin120 \cdot \sin 10}{\sin 20 \cdot \sin 140}\right)^2+1^2-2\left(\dfrac{\sin 120 \cdot \sin 10}{\sin 20 \cdot \sin 140}\right)(1)(\cos 70)$

$AB=1$

Apply sine law on $\triangle ADB$ .

$\dfrac{\sin \angle ABD}{\dfrac{\sin 30}{\sin 140}}=\dfrac{\sin 100}{1}$

$\boxed{\angle ABD=50^\circ}$