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° + ∠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°
APPEARS IN
संबंधित प्रश्न
Find the length of tangent drawn to a circle with radius 8 cm form a point 17 cm away from the center of the circle
In the given figure, the chord AB of the larger of the two concentric circles, with center O, touches the smaller circle at C. Prove that AC = CB.

Two circles touch internally. The sum of their areas is 116 π cm2 and the distance between their centres is 6 cm. Find the radii of the circles ?
A chord of a circle of radius 14 cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle. `[User pi22/7 and sqrt3=1.73]`
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ∠BAO =
If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than ∠AOB =
If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
Draw circle with the radii given below.
4 cm
If two chords AB and CD of a circle AYDZBWCX intersect at right angles (see figure), prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle.

In the following figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

