Advertisements
Advertisements
Question
In the given figure, AB is the diameter of a circle with centre O.
∠BCD = 130°. Find:
- ∠DAB
- ∠DBA
Solution
i. ∠DAB + ∠BCD = 180° ...(Opposite angle of a cyclic quadrilateral)
∠DAB + 130° = 180°
∠DAB = 180° – 130°
∠DAB = 50°
ii. ∠ADB = 90° ...(Angle in semicircle)
In ΔADB,
∠DAB + ∠ADB + ∠DBA = 180° ...(Angle sum property)
50° + 90° + ∠DBA = 180°
∠DBA = 180° – 140°
∠DBA = 40°
APPEARS IN
RELATED QUESTIONS
In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of:
- ∠BCD
- ∠BOD
- ∠OBD
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130o. Find:
1) ∠DAB
2) ∠DBA
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E and the given circle at point F, prove that:
- EF = FC
- BF = DF
In the given figure, O is the centre of the circle. The tangents at B and D intersect each other at point P. If AB is parallel to CD and ∠ABC = 55°, find:
- ∠BOD
- ∠BPD
ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AD and BC produced meet at P, show that APB = 60°.
In a cyclic quadrialteral ABCD , if m ∠ A = 3 (m ∠C). Find m ∠ A.
Prove that the angles bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (provided they are not parallel) intersect at right triangle.
