In the given figure, ABCD is a cyclic quadrilateral, OB is the radius, PB is the tangent at point B, and ∠OBC=30°. AOC is a straight line passing through the centre O. Find the value of x.(2 marks)

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Solution

In the given figrue,

∠OBP =90° and ∠OBC =30°

∠CBP =90°- 30°=60°

∠CAB =60°
(∠CAB = ∠CBP, alternate segment of chord))(1 mark)

In ∆OAB,

OA = OB

∠OAB = ∠OBA = 60°

So, ∠OAB + ∠OBA + ∠AOB =180°

(Sum of angles of a triangle is 180°)(1 mark)

60°+ 60°+ ∠AOB =180°

∠AOB =180°-120° = 60°

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