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प्रश्न
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.
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उत्तर
OD = OB
∴ ∠ODB = ∠OBD
Or ∠ABD = 30°
Also, AB || ED
∴ ∠DBC = ∠ODB = 30° (Alternate angles)
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
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, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°.
Calculate:
- ∠EBA,
- ∠BCD.
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.
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, 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 following figure, AD is the diameter of the circle with centre O. chords AB, BC and CD are equal. If ∠DEF = 110°, Calculate: ∠FAB.
In the given figure, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.
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
