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
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°
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
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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