Question

In a circle with centre O chords AB=BC=CD. Point P is on the circumference such that angle ∠APD=75°.Calculate angle ∠ABC.

Answer 1

Solution :-

→ ∠APD=75°. (given)

→ ∠AOD=75° * 2 = 150° . (angle subtended by chord at the centre is double of angle at circumference by chord AD.)

So,

→ ∠BOA = 180° - 150° = 30° (BD is a straight line.)

Now, in ∆AOB,

→ OA = OB = (radius of circle).

therefore,

→ ∠OAB = ∠OBA (angle opposite to equal sides are equal.)

hence,

→ ∠OBA + ∠OAB+ ∠BOA = 180° (angle sum property.)

→ 2∠OBA + 30° = 180°

→ 2∠OBA = 180° - 30°

→ ∠OBA = (150/2) = 75° ------------- (1)

Now, in ∆BOC,

→ OC = OB = (radius of circle).

therefore,

→ ∠OCB = ∠OBC (angle opposite to equal sides are equal).

and,

→ ∠BOC = 150° (Vertically opposite angle.)

hence,

→ ∠OBC + ∠OCB+ ∠BOC = 180° (angle sum property.)

→ 2∠OBC + 150° = 180°

→ 2∠OBC = 180° - 150°

→ ∠OBC = (30/2) = 15° ------------- (2)

from (1) and (2) , we get,

→ ∠ABC = ∠OBA + ∠OBC

→ ∠ABC = 75° + 15°

→ ∠ABC = 90° .(Ans.)

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