Draw a circle with centre O having radius 3 cm. Draw tangent segments PA and PB through the point P outside the circle such that ∠APB = 70º.

Answer: Construction steps; the angle between the two tangents is 70º because ∠AOB = 110º.

Since the radii OA and OB are perpendicular to the tangents PA and PB, in quadrilateral OAPB, ∠OAP = ∠OBP = 90º, so ∠APB = 180º – ∠AOB. For ∠APB = 70º we need ∠AOB = 110º.

Construction :

1. Draw a circle with centre O and radius 3 cm.
2. Draw any radius OA and mark a point B on the circle such that ∠AOB = 110º.
3. Draw a line through A perpendicular to OA and a line through B perpendicular to OB; let them meet at P.
4. PA and PB are the required tangent segments, and ∠APB = 70º.

Check: OP is the bisector of ∠APB, so ∠APO = 35º and PA = \(\dfrac{OA}{\tan 35^\circ} = \dfrac{3}{0.7002} \approx 4.3\) cm.