Question

In the adjoining figure, O is the centre of a circle.If AB=BC and angle AOB=70°,then angle OBC is
(a) 55° (b) 65° (c) 70° (d) 75°
​

Answer 1

Answer:

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options c

70°

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Answer 2

The angle OBC = 55° (Option A)

Given that, AB=BC

From the figure, we can see that, OA =OB =OC (radius of the circle)

Now In ΔOAB and ΔOCB

Side OA = Side OC  (radius of the same circle)

Side AB = Side CB (Given)

Side OB = Side OB (Common)

By SSS congruency rule, ΔOAB ≅ ΔOCB

⇒ ∠AOB = ∠COB (Properties of  congruent triangles)

⇒ ∠COB = 70°

Now, In ΔOCB

OC = OB (Radius of the same circle)

⇒ ∠OCB = ∠OBC (Properties of an isosceles triangle)

Then, In ΔOCB

∠OCB + ∠OBC + ∠BOC = 180 (Sum of all the angles of a triangle)

2∠OBC + 70° = 180

2∠OBC = 110°

∠OBC = 55°