12. In triangle ABC, BM is the altitude to side AC. If angle C = 30 degrees and angle BAM = 45 degrees, find angle ABC.

Solution:

In the given triangle ABC, BM is the altitude to side AC. This means that angle BMA and angle BMC are right angles, i.e., \( \angle BMA = \angle BMC = 90^{\circ} \).

We are given that \( \angle C = 30^{\circ} \) and \( \angle BAM = 45^{\circ} \).

Consider the right-angled triangle BMC:

The sum of angles in a triangle is \( 180^{\circ} \).

So, in \( \triangle BMC \), \( \angle MBC + \angle BMC + \angle C = 180^{\circ} \).

\( \angle MBC + 90^{\circ} + 30^{\circ} = 180^{\circ} \).

\( \angle MBC + 120^{\circ} = 180^{\circ} \).

\( \angle MBC = 180^{\circ} - 120^{\circ} = 60^{\circ} \).

Now, consider the right-angled triangle BMA:

In \( \triangle BMA \), \( \angle MBA + \angle BMA + \angle BAM = 180^{\circ} \).

\( \angle MBA + 90^{\circ} + 45^{\circ} = 180^{\circ} \).

\( \angle MBA + 135^{\circ} = 180^{\circ} \).

\( \angle MBA = 180^{\circ} - 135^{\circ} = 45^{\circ} \).

The angle ABC is the sum of angles MBA and MBC:

\( \angle ABC = \angle MBA + \angle MBC \).

\( \angle ABC = 45^{\circ} + 60^{\circ} \).

\( \angle ABC = 105^{\circ} \).

Ответ: \( 105^{\circ} \).