Advertisements
Advertisements
प्रश्न
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°. Calculate : ∠BCD.
Advertisements
उत्तर
AB || ED
Therefore ∠DEB = EBA = 27° (Alternate angles)
Therefore BCDE is a cyclic quadrilateral
Therefore ∠DEB + ∠BCD = 180°
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
Therefore ∠BCD = 180° – 27° = 153°
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, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate:
- ∠DAB,
- ∠DBA,
- ∠DBC,
- ∠ADC.
Also, show that the ΔAOD is an equilateral triangle.
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: ∠AEF
In the given figure, AB is the diameter of a circle with centre O.
If chord AC = chord AD, prove that:
- arc BC = arc DB
- AB is bisector of ∠CAD.
Further, if the length of arc AC is twice the length of arc BC, find:
- ∠BAC
- ∠ABC
In the figure, given below, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of lengths 24 cm and 18 cm respectively.
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, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.
In Fig, Chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, Calculate ∠ DEC.
In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
In the given figure, AC is the diameter of the circle with center O.
CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°
Calculate:
- ∠BEC
- ∠BCD
- ∠CED
