Аналогично предыдущей задаче, \(\angle AKO = \angle BKO = \frac{42^{\circ}}{2} = 21^{\circ}\).
\(\angle OAK = 90^{\circ}\).
\(\angle AOK = 180^{\circ} - 90^{\circ} - 21^{\circ} = 69^{\circ}\).
\(\angle AOB = 2 \cdot 69^{\circ} = 138^{\circ}\).
\(\angle OAB = \angle OBA = \frac{180^{\circ} - 138^{\circ}}{2} = \frac{42^{\circ}}{2} = 21^{\circ}\).
Ответ: 21