7.5 In triangle ABC, the median BM is drawn. It is known that $\angle \mathrm{BAC}=30^{\circ}, \angle \mathrm{BMC}=45^{\circ}$. Find angle BAC.
Solution. See fig.

![](https://cdn.mathpix.com/cropped/2024_05_06_99acf95ffb20a6f96cefg-2.jpg?height=471&width=643&top_left_y=695&top_left_x=341)

C

$\triangle \mathrm{AMB}: \angle \mathrm{AMB}=135^{\circ} \Rightarrow \angle \mathrm{ABM}=15^{\circ}$

Let $\mathrm{AC}=2 \mathrm{~b}$. Then $\mathrm{AM}=\mathrm{MC}=\mathrm{b}, \mathrm{CH}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \cdot 2 \mathrm{~b}=\mathrm{b}$ (as the leg opposite the $30^{\circ}$ angle in $\left.\triangle \mathrm{AHC}\right)$.

$\mathrm{MH}=\mathrm{b}$ (as the median from the vertex of the right angle in $\triangle \mathrm{AHC}$).

Thus, $\triangle \mathrm{MHC}$ is equilateral, so

$$
\angle \mathrm{HMC}=60^{\circ} \Rightarrow \angle \mathrm{HMB}=\angle \mathrm{HMC}-\angle \mathrm{BMC}=60^{\circ}-45^{\circ}=15^{\circ} .
$$

Therefore, $\triangle \mathrm{MHB}$ is isosceles: $\mathrm{BH}=\mathrm{MH}=\mathrm{b}$. We get: $\triangle \mathrm{BHC}$ is an isosceles right triangle, $\angle \mathrm{ABC}=\angle \mathrm{HBC}=45^{\circ}$.