Дано: \triangle NOB, \angle 1 = \angle 2, \angle 3 = \angle 4, \angle BHO = 98^{\circ}, OH \perp NB. Найти: углы \triangle NOB.

1. В \triangle OHB, \angle OHB = 90^{\circ}, \angle 1 = \angle 2. Следовательно, \triangle OHB - равнобедренный, OH = HB.

2. В \triangle OНB, \angle HOB = 180^{\circ} - 90^{\circ} - \angle 2 = 90^{\circ} - \angle 2.

3. В \triangle NOB, \angle 3 = \angle 4. Следовательно, ON = NB.

4. В \triangle OHB, \angle 3 = \angle 4. \angle HOB = 90^{\circ} - \angle 2. \angle NOB = \angle 3 + \angle HOB = \angle 3 + 90^{\circ} - \angle 2.

5. В \triangle NOB, \angle O = 98^{\circ}. \angle 3 + \angle 4 + \angle 1 + \angle 2 + \angle O = 180^{\circ}. 2\angle 3 + 2\angle 2 + 98^{\circ} = 180^{\circ}. 2(\angle 3 + \angle 2) = 82^{\circ}. \angle 3 + \angle 2 = 41^{\circ}. \angle 2 = 41^{\circ} - \angle 3.

6. \angle HOB = 90^{\circ} - (41^{\circ} - \angle 3) = 49^{\circ} + \angle 3.

7. \angle NOB = \angle 3 + 49^{\circ} + \angle 3 = 2\angle 3 + 49^{\circ}. \angle NOB = 98^{\circ}. 2\angle 3 + 49^{\circ} = 98^{\circ}. 2\angle 3 = 49^{\circ}. \angle 3 = 24.5^{\circ}. \angle 4 = 24.5^{\circ}. \angle 2 = 41^{\circ} - 24.5^{\circ} = 16.5^{\circ}. \angle 1 = 16.5^{\circ}. \angle N = 180^{\circ} - 98^{\circ} - (24.5^{\circ} + 16.5^{\circ}) = 180^{\circ} - 98^{\circ} - 41^{\circ} = 41^{\circ}. \angle B = \angle 1 + \angle 2 = 16.5^{\circ} + 16.5^{\circ} = 33^{\circ}. \angle O = 98^{\circ}. \angle N = 41^{\circ}. \angle B = 33^{\circ}. 98^{\circ} + 41^{\circ} + 33^{\circ} = 172^{\circ}.

8. В \triangle OHB, \angle HOB = 90^{\circ} - 16.5^{\circ} = 73.5^{\circ}. \angle NOB = \angle 3 + \angle HOB = 24.5^{\circ} + 73.5^{\circ} = 98^{\circ}.

9. \angle N = 180^{\circ} - 98^{\circ} - 33^{\circ} = 49^{\circ}.

Ответ: \angle N = 49^{\circ}, \angle B = 33^{\circ}, \angle O = 98^{\circ}.