11.3. Given a triangle $A B C$. It is known that $\angle B=60^{\circ}, \angle C=75^{\circ}$. On side $B C$ as the hypotenuse, an isosceles right triangle $B D C$ is constructed inside triangle $A B C$. What is the measure of angle $D A C$?
Answer: $30^{0}$.

Solution 1: From the problem statement, it follows that $\angle A=45^{0}$. Draw a circle with center $\mathrm{M}$ and radius $\mathrm{MB}=\mathrm{MC}$. Since $\angle \mathrm{BDC}=90^{\circ}$, the major arc BC is seen at an angle of $45^{0}$. Therefore, vertex A lies on this circle. This means that triangle ADC is isosceles, so $\angle \mathrm{DAC}=\angle \mathrm{DCA}=\angle \mathrm{BCA}-\angle \mathrm{DCB}=75^{\circ}-45^{\circ}=$ $30^{0}$.

Solution 2: Let $\mathrm{BC}=\mathrm{a}$, then from triangle ВМС: $D C=\frac{a}{\sqrt{2}}$. From triangle АВС using the sine rule, we get $\frac{B C}{\sin 45^{\circ}}=\frac{A C}{\sin 60^{\circ}}$, which means $A C=a \sqrt{\frac{3}{2}}$. Next, from triangle CDA using the cosine rule:

$$
A D^{2}=C D^{2}+C A^{2}-2 C D \cdot C A \cdot \cos D C A=\frac{a^{2}}{2}+\frac{3 a^{2}}{2}-2 \cdot \frac{a}{\sqrt{2}} \cdot a \cdot \sqrt{\frac{3}{2}} \cdot \frac{\sqrt{3}}{2}=\frac{a^{2}}{2}
$$

so $A D=\frac{a}{\sqrt{2}}$. Therefore, triangle ADC is isosceles. Further calculations are as in Solution 1.

## Criteria:

7 points - correct answer and a fully justified solution;

5 points - it is proven that triangle $\mathrm{ABC}$ is isosceles, but an arithmetic error is made in further calculations;

only answer - 0 points.