Question 1.
In the figure, O is the centre of a circle passing through points A, B, C and D and ∠ADC = 120°. Find the value of x.

Solution:
Since ABCD is a cyclic quadrilateral
∠ADC + ∠ABC = 180°
[∴ opp. ∠s of a cyclic quad. are supplementary]
120° + ∠ABC = 180°
∠ABC = 180° – 120° = 60°
Now, ∠ACB = 90° [angle in a semicircle]
In rt. ∠ed ∆CB, ∠ACB = 90°
∠CAB + ∠ABC = 90°
x + 60° = 90°
x = 90° -60°
x = 30°

Question 2.
In the given figure, O is the centre of the circle, ∠AOB = 60° and CDB = 90°. Find ∠OBC.

Solution:
Since angle subtended at the centre by an arc is double the angle
subtended at the remaining part of the circle.
∴ ∠ACB = 13 ∠AOB = 13 x 60° = 30°
Now, in ACBD, by using angle sum property, we have
∠CBD + ∠BDC + ∠DCB = 180°
∠CBO + 90° + ∠ACB = 180°
[∵ ∠CBO = ∠CBD and ∠ACB = ∠DCB are the same ∠s]
∠CBO + 90° + 30° = 180°
∠CBO = 180o – 90° – 30° = 60°
or ∠OBC = 60°

Question 3.
In the given figure, O is the centre of the circle with chords AP and BP being produced to R and Q respectively. If ∠QPR = 35°, find the measure of ∠AOB.

Solution:
∠APB = ∠RPQ = 35° [vert. opp. ∠s]
Now, ∠AOB and ∠APB are angles subtended by an arc AB at centre and at the remaining part of the circle.
∴ ∠AOB = 2∠APB = 2 × 35° = 70°

Question 4.
In the figure, PQRS is a cyclic quadrilateral. Find the value of x.

Solution:
In ∆PRS, by using angle sum property, we have
∠PSR + ∠SRP + ∠RPS = 180°
∠PSR + 50° + 35o = 180°
∠PSR = 180° – 85o = 95°
Since PQRS is a cyclic quadrilateral
∴ ∠PSR + ∠PQR = 180°
[∵ opp. ∠s of a cyclic quad. are supplementary]
95° + x = 180°
x = 180° – 95°
x = 85°

Question 5.
In the given figure, ∠ACP = 40° and BPD = 120°, then find ∠CBD.

Solution:
∠BDP = ∠ACP = 40° [angle in same segment]
Now, in ∆BPD, we have
∠PBD + ∠BPD + ∠BDP = 180°
⇒ ∠PBD + 120° + 40° = 180°
⇒ ∠PBD = 180° – 160o = 20°
or ∠CBD = 20°

Question 6.
In the given figure, if ∠BEC = 120°, ∠DCE = 25°, then find ∠BAC.

Solution:
∠BEC is exterior angle of ∆CDE.
∴ ∠CDE + ∠DCE = ∠BEC
⇒ ∠CDE + 25° = 120°
⇒ ∠CDE = 95°
Now, ∠BAC = ∠CDE [∵ angle in same segment are equal]
⇒ ∠BAC = 95°

Question 7.
In the given figure, PQR = 100°, where P, Q and R are points on a circle with centre O. Find LOPR.

Solution:
Take any point A on the circumcircle of the circle.
Join AP and AR.
∵ APQR is a cyclic quadrilateral.
∴ ∠PAR + ∠PQR = 180° [sum of opposite angles of a cyclic quad. is 180°]
∠PAR + 100° = 180°
⇒ Since ∠POR and ∠PAR are the angles subtended by an arc PR at the centre of the circle and circumcircle of the circle.
∠POR = 2∠PAR = 2 x 80° = 160°
∴ In APOR, we have OP = OR [radii of same circle]
∠OPR = ∠ORP [angles opposite to equal sides]
Now, ∠POR + ∠OPR + ∠ORP = 180°
⇒ 160° + ∠OPR + ∠OPR = 180°
⇒ 2∠OPR = 20°
⇒ ∠OPR = 10°

Question 8.
In figure, ABCD is a cyclic quadrilateral in which AB is extended to F and BE || DC. If ∠FBE = 20° and DAB = 95°, then find ∠ADC.

Solution:
Sum of opposite angles of a cyclic quadrilateral is 180°
∴ ∠DAB + ∠BCD = 180°
⇒ 95° + ∠BCD = 180°
⇒ ∠BCD = 180° – 95° = 85°
∵ BE || DC
∴ ∠CBE = ∠BCD = 85°[alternate interior angles]
∴ ∠CBF = CBE + ∠FBE = 85° + 20° = 105°
Now, ∠ABC + 2CBF = 180° [linear pair]
and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad.]
Thus, ∠ABC + ∠ADC = ∠ABC + 2CBF
⇒ ∠ADC = CBF
⇒ ∠ADC = 105° [∵ CBF = 105°]

Question 9
Equal chords of a circle subtends equal angles at the centre.

Solution:
Given : In a circle C(O, r), chord AB = chord CD
To Prove : ∠AOB = ∠COD.
Proof : In ∆AOB and ∆COD
AO = CO (radii of same circle]
BO = DO [radii of same circle]
Chord AB = Chord CD (given]
⇒ ∆AOB = ACOD [by SSS congruence axiom]
⇒ ∠AOB = COD (c.p.c.t.]

Question 10.
In the given figure, P is the centre of the circle. Prove that : ∠XPZ = 2(∠X∠Y + ∠YXZ).

Solution:
Arc XY subtends ∠XPY at the centre P and ∠XZY in the remaining part of the circle.
∴ ∠XPY = 2 (∠X∠Y)
Similarly, arc YZ subtends ∠YPZ at the centre P and ∠YXZ in the remaining part of the circle.
∴ ∠YPZ = 2(∠YXZ) ….(ii)
Adding (i) and (ii), we have
∠XPY + ∠YPZ = 2 (∠XZY + ∠YXZ)
∠XP2 = 2 (∠XZY + ∠YXZ)

Important Link

Quick Revision Notes : Circles

NCERT Solution : Circles

MCQs: Circles

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