
Find the length of PQ:
We are given that \( AP = AQ = 30 \, \text{cm} \), and the angle between the tangents, \( \angle PAQ = 60^\circ \).
To find \( PQ \), we can use the law of cosines in triangle \( PAQ \), where:
\[ PQ^2 = AP^2 + AQ^2 - 2 \times AP \times AQ \times \cos(\angle PAQ) \]
Substituting the given values:
\[ PQ^2 = 30^2 + 30^2 - 2 \times 30 \times 30 \times \cos(60^\circ) \]
Since \( \cos(60^\circ) = 0.5 \):
\(PQ^2 = 900 + 900 - 2 \times 30 \times 30 \times 0.5\)
\(PQ^2 = 900 + 900 - 900 = 900\)
\(PQ = \sqrt{900} = 30 \, \text{cm}\)
Thus, the length of \( PQ \) is \( 30 \, \text{cm} \).
Find \( m \angle POQ \):
Since \( AP \) and \( AQ \) are tangents to the circle from the point \( A \), the angle between the tangents at \( P \) and \( Q \) is equal to the angle at the center of the circle subtended by the chord \( PQ \). This means that:
\[ \angle POQ = 2 \times \angle PAQ \]
Substituting the given value of \( \angle PAQ \):
\[ \angle POQ = 2 \times 60^\circ = 120^\circ \]
Thus, \( m \angle POQ = 120^\circ \).

In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity: 
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
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