Advertisements
Advertisements
प्रश्न
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.

Advertisements
उत्तर

Join AC and BD
∴ ∠CAD = 90° and ∠CBD = 90°
(Angle in a semicircle is a right angle)
Also, AB || CD
∴ ∠BAD = ∠ADC = 25° ...(Alternate angles)
∠BAC = ∠BAD + ∠CAD
= 25° + 90°
= 115°
∴ ∠ADB = 180° – 25° – ∠BAC
= 180° – 25° – 115°
= 40°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
Also, ∠AEB = ∠ADB = 40°
(Angle subtended by the same chord on the circle are equal)
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC

Prove that the rhombus, inscribed in a circle, is a square.
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.

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. 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 figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC

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

