Draw a circle with centre O and radius 3.4 cm. Draw a chord AB of length 6 cm. Draw tangents at A and B and let them meet at P. Measure \(\angle APB\) and \(\angle AOB\). What do you observe?

Answer: \(\angle AOB \approx 124^\circ\) and \(\angle APB \approx 56^\circ\); they add up to 180°, and PA = PB.

Steps of construction:

1. Draw a circle with centre O, radius 3.4 cm. Take point A on it; with A as centre, radius 6 cm, cut the circle at B. Draw chord AB.
2. Draw radii OA and OB. At A draw a line perpendicular to OA and at B a line perpendicular to OB. Let them meet at P.
3. Measure \(\angle AOB\), \(\angle APB\), PA and PB.

Expected values: half of \(\angle AOB\) satisfies \(\sin \dfrac{\angle AOB}{2} = \dfrac{3}{3.4}\), which gives \(\angle AOB \approx 124^\circ\) and so \(\angle APB \approx 56^\circ\).

Observation: \(\angle APB + \angle AOB = 180^\circ\), because in \(\square OAPB\) the angles at A and B are 90° each. Also PA = PB (tangent segments from an external point).

[Figure: circle with centre O; chord AB = 6 cm; radii OA, OB; tangents at A and B meeting at P outside the circle.]