31. Find \(\angle ADC\).

In triangle ABC, \(\angle A = 20^\circ\), \(\angle B = 130^\circ\), so \(\angle C = 180^\circ - (20^\circ + 130^\circ) = 180^\circ - 150^\circ = 30^\circ\).
We are given \(\angle ACB = 15^\circ\). Therefore \(\angle ACD = 30 - 15 = 15^\circ\).
In triangle BCD, \(\angle DBC = 130^\circ - \angle ABD\), \(\angle BCD = 15^\circ\) and \(\angle CDB = 180 -130 =50^\circ \).
Since \(\angle DBC + \angle BCD + \angle BDC = 180^\circ \), \(\angle BDC =180-(15+50)=115\)
Consider triangle ADC, the angles must add to \(180\), so, \(\angle ADC = 180 - (20+15)=145^\circ\)

**Answer: \(\angle ADC = 145^\circ\)**