Advertisements
Advertisements
प्रश्न
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
उत्तर
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° + ∠ABC = 180°
⇒ ∠ABC = 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 = ∠OBC + ∠OBE
= 50° + 50°
= 100°
APPEARS IN
संबंधित प्रश्न
Fill in the blanks:
An arc is a __________ when its ends are the ends of a diameter.
Find the length of a tangent drawn to a circle with radius 5cm, from a point 13 cm from the center of the circle.
In fig. there are two concentric circles with Centre O of radii 5cm and 3cm. From an
external point P, tangents PA and PB are drawn to these circles if AP = 12cm, find the
tangent length of BP.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points?
In Fig., chords AB and CD of the circle intersect at O. AO = 5 cm, BO = 3 cm and CO = 2.5 cm. Determine the length of DO.

A chord is at a distance of 15 cm from the centre of the circle of radius 25 cm. The length of the chord is
Given: A circle inscribed in a right angled ΔABC. If ∠ACB = 90° and the radius of the circle is r.
To prove: 2r = a + b – c

The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel to BC.
A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.
