Mathematics — CBSE Class 10 board question

Construction Steps:

Since the two tangents are inclined at 60°, the angle at the centre ∠POQ = 180° − 60° = 120°.
(Since OP ⊥ PA and OQ ⊥ QA, the angles in quadrilateral OPAQ add to 360°.)

Steps:

1. Draw a circle of radius 4 cm with centre O.
2. Draw any radius OP.
3. Draw another radius OQ such that ∠POQ = 120°.
4. At P, draw a perpendicular to OP (i.e., construct a 90° angle at P).
5. At Q, draw a perpendicular to OQ (i.e., construct a 90° angle at Q).
6. Let the two perpendiculars meet at point A.

PA and QA are the required pair of tangents, inclined to each other at 60°.

Source: Chapter 10, Circles

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• The key relation is: angle between tangents + angle at centre = 180°. So if tangents meet at 60°, ∠POQ = 120°.
• Tangent ⊥ radius at point of contact (Theorem 10.1), so you draw perpendiculars at P and Q.
• Examiners expect the reasoning for ∠POQ = 120° and clear, numbered steps. Marks are given for correct angle calculation, accurate construction steps, and a neat labelled diagram.