Advertisements
Advertisements
Question
In the following figure, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.
Solution
Let us consider, the points A, B, C and D form a cyclic quadrilateral.
∴ ∠ADC + ∠OBC = 180° ...[The sum of opposite angles of a cyclic quadrilateral is 180°]
`\implies` 130° + ∠OBC = 180°
`\implies` ∠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]
`\implies` ∠OBC = ∠OBE ...[By C.P.C.T]
Now, ∠CBE = ∠CBO + ∠EBO
= 50° + 50°
= 100°
APPEARS IN
RELATED QUESTIONS
In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.
In the given figure, PQ and RS are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects PQ at A and RS at B. Prove that ∠AOB = 90º
In Fig below, PQ is tangent at point R of the circle with center O. If ∠TRQ = 30°. Find
∠PRS.
The greatest chord of a circle is called its
If \[d_1 , d_2 ( d_2 > d_1 )\] be the diameters of two concentric circle s and c be the length of a chord of a circle which is tangent to the other circle , prove that\[{d_2}^2 = c^2 + {d_1}^2\].
Find the area of a circle of radius 7 cm.
Use the figure given below to fill in the blank:
Tangent to a circle is _______.
Construct a triangle PQR with QR = 5.5 cm, ∠Q = 60° and angle R = 45°. Construct the circumcircle cif the triangle PQR.
In the adjoining figure, Δ ABC is circumscribing a circle. Then, the length of BC is ______
In the following figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to ______.
