In the adjoining figure 'O' is the center of the circle and PQ, PR an

1.	In the adjoining figure 'O' is the center of the circle and PQ, PR and ST are the three tangents. ∠QPR = 50°, then ∠SOT equals (a) 35° (b) 65° (c) 45° (d) 50°
Answer» Answer : (b) 65º ∠ROQ = 180° – 50° = 130° (∵ ∠OQP + ∠ORP + ∠OPR + ∠ROQ = 360° and ∠OQP = ∠ORP = 90°) RT = TM, QS = SM (Tangents to a circle from the same external point are equal) Also, OQ = OM = OR (Radii of the given circle) ∴ ∠ROT = ∠TOM and ∠MOS = ∠SOQ. (∵ Tangents from the an external point subtend equal angles at the centre) ⇒ ∠SOT = ∠SOM + ∠TOM = \(\frac{1}{2}\) ∠QOM + \(\frac{1}{2}\) ∠ROM ∠SOT = \(\frac{1}{2}\) ∠ROQ = \(\frac{1}{2}\) × 130° = 65°.

In the adjoining figure 'O' is the center of the circle and PQ, PR and ST are the three tangents. ∠QPR = 50°, then ∠SOT equals (a) 35° (b) 65° (c) 45° (d) 50°

Answer»

Answer : (b) 65º

∠ROQ = 180° – 50° = 130° (∵ ∠OQP + ∠ORP + ∠OPR + ∠ROQ = 360° and ∠OQP = ∠ORP = 90°)

RT = TM, QS = SM (Tangents to a circle from the same external point are equal)

Also, OQ = OM = OR (Radii of the given circle)

∴ ∠ROT = ∠TOM and ∠MOS = ∠SOQ. (∵ Tangents from the an external point subtend equal angles at the centre)

⇒ ∠SOT = ∠SOM + ∠TOM = \(\frac{1}{2}\) ∠QOM + \(\frac{1}{2}\) ∠ROM

∠SOT = \(\frac{1}{2}\) ∠ROQ

= \(\frac{1}{2}\) × 130°

= 65°.

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In the adjoining figure 'O' is the center of the circle and PQ, PR and ST are the three tangents. ∠QPR = 50°, then ∠SOT equals (a) 35° (b) 65° (c) 45° (d) 50°

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