Two circles with centers A and B intersect at points P and Q. Given radius of circle A, AP=AQ=10cm, radius of circle B, BP=BQ=4cm, central angle in circle A, ∠PAQ=52°, and central angle in circle B, ∠

The problem describes two circles with centers A and B, intersecting at points P and Q.
Given:
Radius of circle A, $AP = AQ = 10 \text{ cm}$.
Radius of circle B, $BP = BQ = 4 \text{ cm}$.
Central angle in circle A, $\angle PAQ = 52^\circ$.
Central angle in circle B, $\angle PBQ = 97.2^\circ$.

The line segment AB connects the centers of the circles. The line segment PQ is the common chord. The line AB is the perpendicular bisector of PQ. Let M be the point where AB intersects PQ. Thus, $\triangle APM$ and $\triangle