7.3. Given an equilateral triangle $\mathrm{ABC}$. On the sides $\mathrm{AB}$ and $\mathrm{BC}$, isosceles right triangles ABP and BCQ are constructed externally with right angles $\angle \mathrm{ABP}$ and $\angle \mathrm{BCQ}$. Find the angle $\angle \mathrm{PAQ}$.
Answer: 90.

Solution: Consider triangle ACQ. It is isosceles because $\mathrm{AC}=\mathrm{BC}$ (by the condition of the equilateral triangle $\mathrm{ABC}$), and $\mathrm{BC}=\mathrm{CQ}$ (by the condition of the isosceles triangle BCQ). The angle at the vertex $\angle A C Q=60 \circ+90 \circ=150$ . Then $\angle \mathrm{CAQ}=(180 \circ-150 \circ): 2=15$ degrees. The required angle $\angle \mathrm{PAQ}$ $=\angle \mathrm{PAB}+\angle \mathrm{BAQ}=\angle \mathrm{PAB}+\angle \mathrm{BAC}-\angle \mathrm{CAQ}=45 \circ+60 \circ-$ $15 \circ=90 \circ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a46b30dcdeca657878e4g-1.jpg?height=540&width=854&top_left_y=1969&top_left_x=1115)

Criteria: Only the answer - 0 points. Partially correct progress in the solution - 3-5 points. 7 points - correct answer with a correct solution.