pls answer my question....
Answers
In Δ BDC ,
BD = DC [ Given ]
Hence ∠ DBC = ∠BCD [ base ∠s of an isosceles Δ ]
∠BCD = 30° [ same as ∠ DBC ]
∠ BCD + ∠DBC + ∠BDC = 180 ° [ ∠ sum property of Δ ]
==> 30 ° + 30° + ∠BDC = 180°
==> ∠BDC = 180° - ( 30° + 30° )
==> ∠BDC = 180° - 60°
==> ∠BDC = 120°
Hence :
∠BAC + ∠BDC = 180° [ Cyclic quadrilateral ]
==> ∠BAC + 120° = 180°
==> ∠BAC = 180° - 120°
==> ∠BAC = 60°
∠BOC = 2 ∠BAC
[ ∠ subtended at the centre is twice the ∠ at the periphery ]
∠BOC = 2 × 60°
==> ∠BOC = 120°
OB = OC
Hence :
∠OBC = ∠OCB [ base ∠s of an isosceles Δ ]
∠OBC + ∠OCB + ∠BOC = 180° [ ∠ sum property ]
==> ∠OBC + ∠OBC + 120° = 180° [ equal ∠ s ]
==> 2 ∠ OBC = 180° - 120°
==> 2 ∠ OBC = 60°
==> ∠ OBC = 60°/2
==> ∠ OBC = 30°
==> ∠OCB = ∠OBC = 30° [ equal ∠s ]
Now , lets join AO .
∠OAB = ∠OAC [ ∠s subtended by equal chords ( radius ) ]
∠OAB = ∠BAC / 2 [ ∠ s are equal ]
= 60° / 2
= 30°
∠ABO = ∠OAB = 30° [ base ∠ s of isosceles Δ ]
∠ABC = ∠OBA + ∠OBC
==> ∠ABC = 30° + 30°
==> ∠ABC = 60°
The value of ∠ABC is 60°
Hope it helps ;-)
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Answer:
Measure of angle ABC is 60°
Step-by-step explanation:
It is given that the length of BD and DC is equal. So,
∠DBC = ∠DCB = 30°
Now, by the property of sum of angles of triangle,
∠BDC + 30° + 30° = 180°
∠BDC = 120°
As the quadrilateral ABDC is a cyclic quadrilateral,
∠BAC + ∠BDC = 180°
∠BAC = 180 - 120° [ ∵ ∠BDC = 120° ]
∠BAC = 60°
By the theorems of circle,
∠BOC = 2 ∠BAC
∠BOC = 2 x 60°
∠BOC = 120°
Now, joining A and O, so : -
∠OAB = 1 / 2 x ∠BAC
∠OAB = 1 / 2 x 60°
∠OAB = 30°
And, Reflex ∠BOC = 360° - 120°
Reflex ∠BOC = 240°
So, ∠AOB = 1 / 2 x reflex ∠BOC
∠AOB = 1 / 2 x 240°
∠AOB = 120°
By the properties of triangle,
= > sum of all angles = 180°
= > ∠ABO + ∠AOB + ∠OAB = 180°
= > ∠ABO = 180° - 120° - 30°
= > ∠ABO = 30°
In ΔOBC, OC = OB = radius, so ∠OBC = ∠OCB,
∠BOC + ∠OBC + ∠OCB = 180°
∠OBC + ∠OBC = 180 - 120°
2 ∠OBC = 60°
∠OBC = 30°
Therefore,
∠ABC = ∠OBC + ∠ABO
= 30° + 30°
= 60°
Hence,
∠ABC = 60°