
Determine the length of PQ:
Given \( AP = AQ = 30 \, \text{cm} \) and \( \angle PAQ = 60^\circ \).
To find \( PQ \), apply the law of cosines to triangle \( PAQ \):
\[ PQ^2 = AP^2 + AQ^2 - 2 \times AP \times AQ \times \cos(\angle PAQ) \]
Substitute the given values:
\[ PQ^2 = 30^2 + 30^2 - 2 \times 30 \times 30 \times \cos(60^\circ) \]
Using \( \cos(60^\circ) = 0.5 \):
\(PQ^2 = 900 + 900 - 2 \times 30 \times 30 \times 0.5\)
\(PQ^2 = 900 + 900 - 900 = 900\)
\(PQ = \sqrt{900} = 30 \, \text{cm}\)
Therefore, the length of \( PQ \) is \( 30 \, \text{cm} \).
Determine \( m \angle POQ \):
Because \( AP \) and \( AQ \) are tangents to the circle originating from point \( A \), the angle formed by these tangents at \( P \) and \( Q \) is equivalent to the angle at the circle's center subtended by chord \( PQ \). This relationship is expressed as:
\[ \angle POQ = 2 \times \angle PAQ \]
Using the provided value for \( \angle PAQ \):
\[ \angle POQ = 2 \times 60^\circ = 120^\circ \]
Therefore, \( m \angle POQ = 120^\circ \).



