Question

7. In the figure O is the center and AB=BC (a) Prove that AOB= BOC? (b) If OA=AB=BC ,find angle AOB & angle BOC?​

Answer 1

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.

Answer 2

In the figure O is the centre of the circle AB and BC are chords also AB =BC and angle OBA=50 degree. Since AB = BC as chords, and OA = OC being radii of he same circle, OABC is a kite.

In the figure O is the centre of the circle AB and BC are chords also AB =BC and angle OBA=50 degree. Since AB = BC as chords, and OA = OC being radii of he same circle, OABC is a kite.If <OBA = 50 then <OBC is also 50.

In the figure O is the centre of the circle AB and BC are chords also AB =BC and angle OBA=50 degree. Since AB = BC as chords, and OA = OC being radii of he same circle, OABC is a kite.If <OBA = 50 then <OBC is also 50.ABC is an isosceles triangle hence <BAC = <BCA = (180–2*50)/2 = 40 deg.

In the figure O is the centre of the circle AB and BC are chords also AB =BC and angle OBA=50 degree. Since AB = BC as chords, and OA = OC being radii of he same circle, OABC is a kite.If <OBA = 50 then <OBC is also 50.ABC is an isosceles triangle hence <BAC = <BCA = (180–2*50)/2 = 40 deg.Reflex <AOC = 2*<ABC = 2*(50+50) = 200 deg, or <AOC = 360–200 = 160 deg. Hence <AOB = <COB = 160/2 = 80 deg. And <OAC = <OCA = 10 deg.