11.2. In triangle $A B C$, angle $C$ is $75^{\circ}$, and angle $B$ is $60^{\circ}$. The vertex $M$ of the isosceles right triangle $B C M$ with hypotenuse $B C$ is located inside triangle $A B C$. Find angle $M A C$.
Answer: $30^{\circ}$.

First method. From the problem statement, it follows that $\angle B A C=45^{\circ}$. Draw a circle with center $M$ and radius $M B=M C$ (see Fig. 11.2). Since $\angle B M C=90^{\circ}$, the larger arc $B C$ of this circle is the locus of points from which the chord $B C$ is seen at an angle of $45^{\circ}$. Therefore, vertex $A$ lies on this circle. Hence, triangle $A M C$ is isosceles, so $\angle M A C=\angle M C A=\angle B C A-\angle M C B=75^{\circ}-45^{\circ}=30^{\circ}$.

Second method. Let $B C=a$, then from triangle $B M C: M C=\frac{a}{\sqrt{2}}$. From triangle $A B C$ using the Law of Sines, we get $\frac{B C}{\sin 45^{\circ}}=\frac{A C}{\sin 60^{\circ}}$, which means $A C=a \sqrt{\frac{3}{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-1.jpg?height=457&width=421&top_left_y=874&top_left_x=1619)
Next, from triangle $C M A$ using the Law of Cosines: $A M^{2}=C M^{2}+C A^{2}-2 C M \cdot \operatorname{CA} \cdot \cos \angle M C A=\frac{a^{2}}{2}+\frac{3 a^{2}}{2}-2 \frac{a}{\sqrt{2}} \cdot a \times$ $\times \sqrt{\frac{3}{2}} \cdot \frac{\sqrt{3}}{2}=\frac{a^{2}}{2}$, so $A M=\frac{a}{\sqrt{2}}$. Therefore, triangle $A M C$ is isosceles.

Further calculations are outlined above.

## Grading Criteria:

+ correct answer and a fully justified solution

$\pm$ it is proven that triangle $A M C$ is isosceles, but an arithmetic error is made in further calculations

- only the answer is provided