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प्रश्न
In the following figure, AD is the diameter of the circle with centre O. chords AB, BC and CD are equal. If ∠DEF = 110°, Calculate: ∠FAB.
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उत्तर
Join AE , OB and OC
∵ Chord AB = Chord BC = Chord CD ...[given]
∴ ∠AOB = ∠BOC = ∠COD ...(Equal chords subtends equal angles at the centre)
But ∠AOB + ∠BOC + ∠COD = 180° ...[AOD is a straight line ]
∠AOB = ∠BOC = ∠COD = 60°
In ∠OAB, OA = OB
∴ ∠OAB = ∠OBA ...[radii of the same circle]
But ∠OAB + ∠OBA = 180° − AOB
= 180° − 60°
= 120°
∴ ∠OAB = ∠OBA = 60°
In cyclic quadrilateral ADEF,
∠DEF + ∠DAF = 180°
⇒ ∠DAF = 180° − ∠DEF
= 180° − 110°
= 70°
Now, ∠FAB = ∠DAF + ∠OAB
= 70° + 60°
= 130°
संबंधित प्रश्न
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
1) Prove that AC is a diameter of the circle.
2) Find ∠ACB
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠NRM
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°. Calculate : ∠BCD.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBA
Also, show that the ΔAOD is an equilateral triangle.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBC
Also, show that the ΔAOD is an equilateral triangle.
In the given figure, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.
In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
