Math, asked by Hritiknicalson, 1 year ago

In the given figure AB and CD are diameters of a circle with centre O.If angle OBD=40° find angle AOC and angle ACD


Anonymous: where is the figure ?

Answers

Answered by Anonymous
13
Hi there !!

I HOPE this is the diagram which u mentioned.

===========================================================
Given :

∠OBD = 40°

∠ABD is the same as ∠OBD.

Hence ,
∠ABD = 40°

Also ,

∠ABD =∠ACD = 40° [ Angles in the same segment]

Hence ,
∠ACD = 40°
=====================================

Now ,
Consider Δ AOC

AO = CO [ Radii of the circle ]

AS two sides of the triangle are equal . the angles are also equal.

∠OAC =∠OCA [ angles opposite to equal sides of a traingles are also equal]

∠OCA is the same as ∠ACD.


∠OAC =∠OCA = 40°


We know that in a triangle , the angle sum is 180°

In Δ AOC ,
∠OAC +∠OCA + ∠AOC = 180°

40°+40° + ∠AOC = 180°

∠AOC = 180 - 80

= 100°
Attachments:
Answered by kashree
4

Now ,

Consider Δ AOC

AO = CO [ Radii of the circle ]

AS two sides of the triangle are equal . the angles are also equal.

∠OAC =∠OCA [ angles opposite to equal sides of a traingles are also equal

∠OCA is the same as ∠ACD.

∠OAC =∠OCA = 40°

We know that in a triangle , the angle sum is 180°

In Δ AOC

∠OAC +∠OCA + ∠AOC = 180°

40°+40° + ∠AOC = 180°

∠AOC = 180 - 80

= 100°





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