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
In Fig. 1, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is :
(A) 3.8
(B) 7.6
(C) 5.7
(D) 1.9
In the given figure, the incircle of ∆ABC touches the sides BC, CA and AB at D, E, F respectively. Prove that AF + BD + CE = AE + CD + BF = `\frac { 1 }{ 2 } ("perimeter of ∆ABC")`
Two tangent segments PA and PB are drawn to a circle with center O such that ∠APB = 120°. Prove that OP = 2AP
In two concentric circles, a chord of length 8 cm of the large circle touches the smaller circle. If the radius of the larger circle is 5 cm, then find the radius of the smaller circle.
In the given figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.
In the given figure, O is the centre of the circle and ∠BDC = 42°. The measure of ∠ACB is
Construct a triangle PQR with QR = 5.5 cm, ∠Q = 60° and angle R = 45°. Construct the circumcircle cif the triangle PQR.
Construct a triangle ABC with AB = 5 cm, ∠B = 60° and BC = 6. 4 cm. Draw the incircle of the triangle ABC.
In the figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is
A line segment which joins any two points on a circle is a ___________
