6. (7 points) In triangle $A B C$, the median $A M$ is drawn. Find the angle $A M C$, if angles $B A C$ and $B C A$ are equal to $45^{\circ}$ and $30^{\circ}$ respectively.

Answer: $135^{\circ}$.
Solution. Let $B H$ be the height of triangle $A B C$. According to the problem, angle $B A C$ is $45^{\circ}$, so $B H = A H$. In triangle $C B H$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B C = 2 B H$. The median $H M$ of the right triangle $B H C$ is equal to half the hypotenuse $B C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2509cbb2376df96befe7g-3.jpg?height=468&width=711&top_left_y=1539&top_left_x=730)

Combining all the equalities of the segments, we get $A H = B H = H M = M B = M C$. Therefore, triangle $M B H$ is equilateral, and angle $C M H$ is $120^{\circ}$. Additionally, triangle $A H M$ is isosceles, its angle $A H M$ is $90^{\circ} + 60^{\circ} = 150^{\circ}$, so angle $A M H$ is $15^{\circ}$. Thus,

$$
\angle A M C = \angle A M H + \angle H M C = 120^{\circ} + 15^{\circ} = 135^{\circ}.
$$

Criteria. Any correct solution: 7 points.

Proved that triangle $A H M$ is isosceles, but the correct answer was not obtained: 5 points.

Only the correct answer was provided: 1 point.