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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
Prove that the parallelogram, inscribed in a circle, is a rectangle.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
In the figure, given alongside, AB || CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°, calculate:
- ∠DAB,
- ∠BDC.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.
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 figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
