Question

Find the length of $P Q$.

Answer 1

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 \( 30 \, \text{cm} \).

Answer 2

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, \( m \angle POQ = 120^\circ \).

Answer 3

(a) Find the length of OA:

To find the length of OA, we can use the law of cosines in triangle OAP, where O is the center of the circle, P is the point of tangency, and A is the external point.

Since \(\angle OAP = 90^\circ\) (the angle between the radius and the tangent is always \(90^\circ\)), triangle OAP is a right triangle. Using the Pythagorean theorem:

\[ OA^2 = OP^2 + AP^2 \]

We know that OP is the radius of the circle, and AP = 30 cm. Let’s denote the radius by \(r\).

\[ OA^2 = r^2 + 30^2 \]

But, OA is the hypotenuse of the right triangle OAP, and we know that \(\angle PAQ = 60^\circ\), which makes the distance from A to the center O (i.e., OA) double the radius:

\[ OA = 2r \]

Substitute \(OA = 2r\) into the Pythagorean theorem:

\[ (2r)^2 = r^2 + 30^2 \] \[ 4r^2 = r^2 + 900 \] \[ 3r^2 = 900 \] \[ r^2 = 300 \] \[ r = \sqrt{300} = 10\sqrt{3} \approx 17.32 \, \text{cm} \]

Thus, OA is twice the radius:

\[ OA = 2 \times 17.32 = 34.64 \, \text{cm} \]

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(b) Find the radius of the mirror:

The radius \(r\) of the mirror is \(\boxed{17.32 \, \text{cm}}\).