Advertisements
Advertisements
Questions
In the following figure, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.
In the given figure, AB is a diameter of the circle with centre O. If ∠ADC = 130° and chord BC = chord BE, find ∠CBE.
Advertisements
Solution
Let us consider the points A, B, C, and D, which form a cyclic quadrilateral.
∴ ∠ADC + ∠OBC = 180° ... [The sum of opposite angles of a cyclic quadrilateral is 180°.]
⇒ 130° + ∠OBC = 180°
⇒ ∠OBC = 180° – 130° = 50°
In ΔBOC and ΔBOE,
BC = BE ...[Given]
OC = OE ... [Radii of a same circle]
And OB = OB ...[Common]
∴ ΔBOC ≅ ΔBOE ...[By SSS congruency]
⇒ ∠OBC = ∠OBE ...[By C.P.C.T]
Now, ∠CBE = ∠CBO + ∠EBO
= 50° + 50°
= 100°
RELATED QUESTIONS
Prove that the line segment joining the point of contact of two parallel tangents to a circle is a diameter of the circle.
In the given figure, if arc AB = arc CD, then prove that the quadrilateral ABCD is an isosceles– trapezium (O is the centre of the circle).
PQ is a chord of length 4.8 cm of a circle of radius 3cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
In Fig. 1, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then the length of each tangent is:
In the given figure, PO \[\perp\] QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OTare right bisector of each other.
In the given figure, O is the centre of a circle, chord PQ ≅ chord RS If ∠ POR = 70° and (arc RS) = 80°, find –
(1) m(arc PR)
(2) m(arc QS)
(3) m(arc QSR)
Construct a triangle ABC with AB = 4.2 cm, BC = 6 cm and AC = 5cm. Construct the circumcircle of the triangle drawn.
Find the radius of the circle
Diameter = 30 cm
If the angle between two tangents drawn from a point P to a circle of radius ‘a’ and centre ‘O’ is 90°, then OP = ______
In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.
