5. A circle is circumscribed around a right triangle $\mathrm{ABC}$ with hypotenuse $\mathrm{AB}$. On the larger leg $\mathrm{AC}$, a point $\mathrm{P}$ is marked such that $\mathrm{AP}=\mathrm{BC}$. On the arc $\mathrm{ACB}$, its midpoint $\mathrm{M}$ is marked. What can the angle $\mathrm{PMC}$ be equal to?
5. Answer: 90 degrees

Consider triangles АРМ and
ВСМ. In them, ВС=АР by condition,
АМ=ВМ - chords subtending equal
arcs, angles МВС and МАС are equal,
since they subtend the same arc.
Therefore, the triangles are equal.

Thus, angles ВМС and АМР are equal. Since angle АМВ is 90 degrees, then angle $\mathrm{PMC}=\mathrm{AMB}-\mathrm{AMP}+\mathrm{BMC}=90$.

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