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SSC CHSL Previous Paper 96 (Held On: 15 Oct 2020 Shift 1)

Option 2 : 62°

**Given :-**

A circle is with center O

Two tangents AP and AQ are drawn from an external point A.

∠POQ = 118°

**Concept :- **

Length of tangents drawn from an outside point to a circle are equal.

It also makes an quadrilateral.

In a quadrilateral sum of all angles are 360°

**Calculation :-**

⇒

⇒ As OP and OQ is radius

⇒ OP = OQ

⇒ PA = PQ (From circle properties)

⇒ ∠APO = ∠AQO = 90° (As OP and OQ is perpendicular to PA and AQ)

⇒ Sum of angles = ∠P + ∠O + ∠Q + ∠A

⇒ 360° = 90° + 118° + 90° + ∠A

⇒ 360° - (90° + 118° + 90°) = ∠A

⇒ 360° - 298° = ∠A

⇒ ∠A = 62°

⇒ The ∠PAQ = 62°

**∴ ∠PAQ = 62° **

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