Draw a circle with centre O and radius 3.5 cm. Draw tangents PA and PB from an external point P such that the angle between them is 60°. Also find the length of PO and of each tangent segment.

Answer: \(\angle AOB = 120^\circ\); PO = 7 cm and PA = PB = \(3.5\sqrt{3} \approx 6.1\) cm.

Analysis: In \(\square OAPB\), \(\angle A = \angle B = 90^\circ\), so \(\angle AOB = 360^\circ - 180^\circ - 60^\circ = 120^\circ\). OP bisects \(\angle APB\), so \(\angle APO = 30^\circ\). In the 30°–60°–90° triangle OAP, OA (opposite 30°) = 3.5 cm, so PO = 2 × 3.5 = 7 cm and \(PA = \sqrt{3} \times 3.5 \approx 6.06\) cm.

Steps of construction:

1. Draw a circle with centre O, radius 3.5 cm. Draw radius OA.
2. Draw radius OB such that \(\angle AOB = 120^\circ\).
3. Draw lines perpendicular to OA at A and to OB at B; let them meet at P.
4. PA and PB are the required tangents with \(\angle APB = 60^\circ\). Measure PO (7 cm) and PA (about 6.1 cm).

[Figure: circle with centre O, radius 3.5 cm; radii OA, OB at 120°; tangents at A and B meeting at P with \(\angle APB = 60^\circ\); OP = 7 cm.]

Alternative: since OP = 7 cm is known, take P with OP = 7 cm and use the usual “bisect OP” method.