PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that ∠APB = 100°, then ∠OAB is equal to:

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SSC CPO Previous Paper 35 (Held On: 23 November 2020 Shift 2)
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  1. 45°
  2. 70°
  3. 50°
  4. 35°

Answer (Detailed Solution Below)

Option 3 : 50°
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SSC CPO : General Intelligence & Reasoning Sectional Test 1
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Detailed Solution

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Given:

∠APB = 100°

O is the centre of the circle.

PA and PB are two tangents drawn from P outside the circle

Concept used:

tangent to a circle forms a right angle with the circle's radius.

The Sum of all angles of a quadrilateral is 360°.

Calculation:

F1 Shraddha Suraj 09.01.2021 D2

tangent to a circle forms a right angle with the circle's radius

OAP = OBP = 90°

The Sum of all angles of a quadrilateral is 360°

OAP + OBP + APB + AOB = 360°

90° + 90° + 100° + AOB = 360°

AOB = 360° - 280°

AOB = 80°

In ΔAOB

AOB + OAB + OBA = 180°

80° + x + x = 180°

2x = 100°

x = 50°

The value of ∠OAB is 50°.

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