O is the centre of the circle in the diagram. If AB = BC,(a) Then prov

1.	O is the centre of the circle in the diagram. If AB = BC,(a) Then prove that ∠AOB = ∠BOC(b) If OA = AB = BC, then find the values of ∠AOB and ∠BOC?(c) Find out how many equilateral triangles can be drawn in a circle with length of its side is radius.
Answer» (a) OA = OB = OC, AB = BC ∆ OAB and ∆ OBC are equal triangles. ∴ ∠AOB and ∠BOC are equal which are opposite to the equal sides AB and BC. (b) If OA = AB then ∆ OAB is an equilateral triangle. If OB = BC, ∆ OBC is equilateral triangle. ∴ ∠AOB = ∠BOC = 60° (c) Each angle at O is 60°. The angle at the centimeter is O is 360° and 6 triangles can be drawn.

O is the centre of the circle in the diagram. If AB = BC,(a) Then prove that ∠AOB = ∠BOC(b) If OA = AB = BC, then find the values of ∠AOB and ∠BOC?(c) Find out how many equilateral triangles can be drawn in a circle with length of its side is radius.

Answer»

(a) OA = OB = OC, AB = BC

∆ OAB and ∆ OBC are equal triangles.

∴ ∠AOB and ∠BOC are equal which are opposite to the equal sides AB and BC.

(b) If OA = AB then ∆ OAB is an equilateral triangle.

If OB = BC, ∆ OBC is equilateral triangle.

∴ ∠AOB = ∠BOC = 60°

(c) Each angle at O is 60°. The angle at the centimeter is O is 360° and 6 triangles can be drawn.