c11Question 1: In the given figure, O is the center of the circle. \angle OAB=30^o and \angle OCB = 40^o . Find \angle AOC .

Answer:

In \triangle AOC

\angle OAC = \angle  OCA = x^o

c11xTherefore \angle  AOC + 2x^o = 180^o

\Rightarrow \angle  AOC = 180^o-2x^o

In \triangle ABC

30^o + x^o + \angle ABC + 40^o + x^o = 180

\angle ABC = 110^o-2x^o

Therefore \angle ABC = 110^o- (180^o-\angle AOC)

\angle ABC = \angle  AOC = 70^o

We know \angle  AOC = 2 \angle ABC (Theorem 9)

Hence \frac{1}{2} \angle AOC = \angle AOC - 70^o

\frac{1}{2} \angle AOC = 70^o \Rightarrow \angle AOC= 140^o

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c12Question 2: In the given figure, \angle BAD = 65^o, \angle ABD = 70^o and \angle BDC = 45^o

(i) Prove that AC is a diameter of the circle.

(ii) find \angle ACB    [2013]

Answer:

Given \angle BAD = 65^o, \angle ABD = 70^o and \angle BDC = 45^o .

(i) In \triangle ABD

65^o+70^o+ \angle ADB = 180^o

\Rightarrow \angle ADB = 180^o-135^o = 45^o

Therefore \angle ADC = 45^o+45^o = 90^o

Therefore AC is the diameter (Theorem 11)

(ii) \angle ACB = \angle ADB (angles in the same segment)

Therefore \angle ACB = 45^o since \angle ADB = 45^o

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c13Question 3: Given O is the center of the circle and \angle AOB = 70^o . Calculate the value of (i) \angle OCA (ii) \angle OAC

Answer:

OA = OC =   radius

Therefore \angle OCA = \angle OAC = y^o

In \triangle AOB

OA = OB = radius

Therefore \angle OAB = \angle OBA = x^o

\therefore 2x^o+70^o =180 \Rightarrow x = 55^o

\angle COA + 70^o = 180^o \Rightarrow COA = 110^o

In \triangle COA

2y^o+ 110^o = 180^o \Rightarrow y = 35^o

(i) \angle OCA = 35^o

(ii) \angle OAC = 35^o

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Question 4: In each of the following figures, O is the center of the circle. Find the values of a, b \ and \  c .

(i) c141 (ii) c142

Answer:

(i) 2b= 130^o  \Rightarrow b = 65^o

Hence 130^o+ 2a = 360^o \Rightarrow a = 115^o

(ii) 2c = 360^o-112^o \Rightarrow c = 124^o

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Question 5: In each of the following figures, O is the center of the circle. Find the values of a, b, c \ and \  d [2007]

(i)c151 (ii)c152
(iii)c153 (iv)c154

Answer:

(i) BD is the diameter

\therefore \angle DAB = 90^o

\therefore \angle ADB = 180^o-90^o-3^o5=55^o

Since \angle ADB = \angle ACB = 55^o (angle in same segment) Therefore a = 55^o

(ii) \angle ADB = \angle ACB (angle in same segment)

In \triangle ECB ,

\angle BEC = 180^o-120^o = 60^o

\therefore \angle ACB = 180^o - 60^o - 25^o = 95^o

\therefore \angle ADB = 95^o= b

(iii) In \triangle AOB

AO = OB = radius

2 \angle ACB = \angle AOB

\therefore \angle AOB = 100^o

\therefore 2c+ 100^o= 180^o \Rightarrow c = 40^o 

(iv) Since AB is the diameter

\therefore \angle BAP = 180^o-90^o-45^o = 45^o

\angle PAB = \angle PCB = 45^o \Rightarrow d = 45^o (angles in the same segment)

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c16Question 6: In the figure, AB is common chord of the two circles. If AC \ and \  AD are diameters, prove that D, B \ and \  C are in straight line. O_1 \ and \  O_2 are the centers of the two circles.

Answer:

Since AD is the diameter

\angle ABD = 90^o

Also since AC is the diameter

\angle ABC = 90^o

\therefore \angle DBC = 90^o+90^o = 180^o

Therefore DBC is a straight line and hence D, B \ and \  C are collinear.

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c17Question 7: In the figure given below, find (i) \angle BCD (ii)  \angle ADC   (iii) \angle ABC

Answer:

ABCD is a cyclic quadrilateral

\therefore \angle BCD + \angle BAD = 180^o

\Rightarrow \angle BCD = 180^o-105^o = 75^o

Also \angle ADC + \angle ABC = 180^o

\angle ABC = 180^o-75^o = 105^o

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c18Question 8: In the given figure, O is the center of the circle. If  \angle AOB = 140^o and  \angle OAC = 50^o . Find (i)  \angle ACB    (ii)  \angle OBC   (iii)  \angle OAB    (iv) \angle CBA

Answer:

\angle ACB = \frac{1}{2} Reflex(\angle AOB) = \frac{1}{2}(360^o-140^o) = 110^o

c18xSince OA = OB (radius of the same circle)

\angle OBA = \angle OAB = \frac{1}{2}(180^o-140^o) = 20^o

\therefore \angle CAB = 50^o -20^o = 30^o

In \triangle CAB

\angle CBA = 180^o - 110^o - 40^o = 40^o

\therefore \angle OBC = \angle CBA + \angle OBA = 40^o+20^o = 60^o 

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c19Question 9: Calculate (i)  \angle CDB   (ii)  \angle ABC   (iii)  \angle ACB

Answer:

\angle CDB = \angle BAC = 49^o (BC subtends same angle in the same segment)

\angle ABC = \angle ADC = 43^o (AC subtends same angle in the same segment)

\therefore \angle ACB = 180^o -  49^o - 43^o = 88^o

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c110Question 10: In the figure given below, ABCD is a cyclic quadrilateral in which  \angle  BAD = 75^o \angle ABD = 58^o and  \angle ADC = 77^o . Find: (i)  \angle BDC (ii)  \angle BCD (iii)  \angle BCA

Answer:

(i) \angle ADB = 180^o-75^o-58^o = 47^o

c110x\therefore \angle BDC = \angle ADC - \angle ADB

= 77^o-47^o = 30^o

(ii) \angle BAD + \angle BCD = 180^o (cyclic quadrilateral)

(iii) \angle ADB = \angle ACB = 47^o

\therefore \angle BCD = 180^o-75^o = 105^o

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c21Question 11: In the following figure, O is the center and \triangle ABC is equilateral. Find (i) \angle ADB (ii) \angle AEB

Answer:

In \triangle ABC , since it is equilateral, all angles are equal.

\therefore \angle CAB= \angle CBA = \angle ACB = 60^o

c31\angle ACB = \angle ADB (angles int he same segment)

\therefore \angle ADB = 60^o

Since ADBE is a cyclic quadrilateral

\angle ADB+ \angle AEB = 180^o \Rightarrow \angle AEB = 180^o-60^o= 120^o

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c22Question 12: Given: \angle CAB = 75^o and \angle CBA=50^o . Find the value of \angle DAB + \angle ABD .

Answer:

\angle ACB = 180^o-75^o-5^o = 55^o

\angle ADB = 55^o (angles in same segment)

Let \angle BCD = \angle BAD = x^o (angles in same segment)

Let \angle DBC = \angle DAC = y^o = 75^o

\angle CAD = \angle CBD

75^o-x^o=y^o \Rightarrow x^o+y^o = 7^o5

\therefore \angle DAB+\angle ABD = 50^o+(x^o+y^o)= 50^o+75^o = 125^o

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c23Question 13: ABCD is a cyclic quadrilateral in circle with center O . If \angle ADC = 130^o ; find \angle BAC .

Answer:

\angle ADC = 130^o

\therefore \angle ABC = 180^o-130^o=50^o

\angle ACB = 90^o (angle in the semi circle)

\therefore \angle BAC = 180^o-50^o-90^o = 40^o

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c24Question 14: In the given figure, AOB is a diameter of the circle and \angle AOC=110^o . Find \angle BDC .

Answer:

\angle BOC = 180^o-110^o = 70^o

OA = OC (radius of the same circle)

c32Therefore in \triangle AOC

\angle OAC = \angle OCA = 35^o

\therefore \angle BDC = \angle BAC = 35^o (angle in the same segment)

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c25Question 15: In the following figure, O is the center of the circle, \angle AOB=60^o and \angle BDC=100^o . Find \angle OBC .

Answer:

Let \angle OBC = x

\angle BOA = 60^o

Hence \angle BCA = 30^o

\therefore 30+100+x = 180 \Rightarrow x = 50^o

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c26Question 16: ABCD is a cyclic quadrilateral in which \angle DAC=27^o, \angle DBA=50^o and \angle ADB=33^o . Calculate (i) \angle DBC (ii) \angle DCB (iii) \angle CAB

Answer:

\angle ADB = \angle ACB = 30^o

(i) \angle DBA = 180^o-70^o-27^o-33^o = 50  

\angle DCB = 50^o+3^o3 = 83^o

(ii) \angle DAC = \angle DBC = 27^o

ABCD is a cyclic quadrilateral

\angle CAB = 180^o-50^o-27^o-33^o = 70^o

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c27Question 17: In the figure given below, AB is the diameter of the circle whose center is O . Given that: \angle ECD = \angle EDC = 32^o . Show that \angle COF = \angle CEF .

Answer:

\angle CED = 180^o-64^o = 116^o

\angle CEF = 64^o

In\triangle COF

OF = OC (radius of the same circle)

2 \angle FDC = \angle FOC \Rightarrow \angle FOC = 64^o

Hence \angle COF = \angle CEF

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c28Question 18: In the figure given below, AB and CD are straight lines through the center O of the circle. If \angle AOC=80^o and \angle CDE=40^o find (i) \angle DCE (ii) \angle ABC

Answer:

\angle CED = 90^o (angle in a semi circle)

(i) Therefore \angle DCE = 180^o-90^o-40^o = 50^o

\angle COB = 180^o-80^o=100^o

\therefore \angle ABC = 180^o-100^o-50^o = 40^o

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c29Question 19: In the given figure, AC is a diameter of a circle whose center is O . A circle is described on AO as a diameter. AE is a chord of the larger circle intersects the smaller circle at B . Prove AB = BE .

Answer:

\angle ABO = \angle AEC = 90^o (angles in a semi circle)

c33\angle OAB = \angle CAE (common angle)

\therefore \triangle ABO \sim \triangle AEC

Hence \frac{AO}{AC}=\frac{AB}{AE}

\Rightarrow \frac{1}{2} = \frac{AB}{AB+BE}

\Rightarrow AB = BE

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c210Question 20: In the following figure, (i) if \angle BAD=96^o , find \angle BCD and \angle BFE . (ii) Prove that AD is parallel to FE .

Answer:

(i) ABCD is a cyclic quardilateral

\therefore \angle DAC + \angle BCD = 180^o

\Rightarrow \angle BCD = 180^o-90^o=84^o

\therefore \angle BCE = 180^o-84^o=96^o

\angle BCE+\angle BFE = 180^o-96^o=84^o

(ii) Since \angle  BCD = \angle BFE = 84^o

AD \parallel FE

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Question 21: Prove that: (i) the parallelogram inscribed in a circle is a rectangle (ii) the rhombus, inscribed in a circle is a square.

Answer:

(i) Let ABCD be a parallelogram inscribed in the circle.

\therefore \angle BAD = \angle BCD (opposite angles of a parallelogram are equal)

\angle BAD + \angle BCD = 180^o

\Rightarrow \angle BAD = \angle BCD = 90^o

Similarly, \angle ABC = \angle ADC = 90^o

Therefore ABCD is a rectangle

(ii) ABCD is a rhombus (given) i.e. all four sides are equal.

\angle BAD = \angle BCD = 90^o

Similarly, \angle ABC = \angle ADC = 90^o

Therefore ABCD is a square

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c45Question 22: In the following figure AB = AC . Prove that DEBC is an isosceles trapezium.

Answer:

AB = AC (given)

\angle ABC = \angle ACB

\angle CBA+\angle CED = 180^o (BCED is a cyclic quadrialteral)

\angle ACB + \angle CED = 180^o

DE \parallel BC … … … … (i)

\angle ADE = \angle ABC (corresponding angles)

\angle AED = \angle ACB

Hence \angle ADE = \angle AED

\Rightarrow AD = AE

AB-AD = AC-AE

\Rightarrow BD = EC … … … … (ii)

Hence because of (i) and (ii), ABCD is an isosceles trapezium.

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Question 23: Two circles intersect at P and Q . Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

c46Answer:

\angle PQA = 90^o (angle in a semi circle)

\angle PQB = 90^o (angle in a semi circle)

Therefore \angle AQB = 180^o

Therefore A, Q \ and \  B are collinear.

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Question 24: ABCD is a quadrilateral inscribed in a circle. having \angle A = 60^o . O is the center of the circle. Show that: \angle OBD+\angle ODB= \angle CBD+\angle CDB .

c47Answer:

\angle BAD + \angle BCD = 180^o (cyclic quadrilateral)

\angle BCD = 180^o-60^o=120^o

\angle BOD = 2 \angle BAD

\angle CBD + \angle CDB = 180^o-120^o = 60^o (sum of the angles in a triangle is 180)

Similarly \angle OBD + \angle ODB = 180^o-120^o=60^o

Therefore \angle OBD + \angle ODB = \angle CBD + \angle CDB

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c44Question 25: The figure given below shows the circle with center O . Given \angle AOC = a and \angle ABC=b . (i) Find the relationship between a \ and  \ b . (ii) Find \angle OAB if OABC is a parallelogram.

Answer:

(i) \angle AOC = a and \angle CBA = b (given)

\angle ABC = \frac{1}{2} Reflex (\angle COA)

b = \frac{1}{2} (360^o-a)

2b = 360^o-a \ or \ a+2b=360^o

(ii) If OABC is a parallelogram

Therefore a = b (opposite angles are equal)

\therefore 3b = 360^o \Rightarrow b = 120^o

\therefore a = 120^o

\angle ACB = \angle BAC

\therefore \angle BAC = 30^o

OC = OA (radius of the same circle)

\therefore \angle OCA = \angle OAC 

\therefore \angle OAC = 30^o

Hence \angle OAB = 30^o+30^o=60^o

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Question 26: Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD at the center O is equal to twice the \angle APC .

c48Answer:

\angle AOC = 2 \angle ADC

Similarly, \angle BOD = 2 \angle BAD

Adding the two

\angle AOC + \angle BOD = 2(\angle ADC+\angle BAD) … … … … (i)

In  \triangle PAD

180^o-\angle APC + \angle PAD + \angle ADC = 180^o

\therefore \angle PAD + \angle ADC = \angle APC … … … … (ii)

Using (1) and (ii)

\angle AOC + \angle BOD = 2(\angle BAD + \angle ADC) = 2 \angle APC

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c43Question 27: In the given figure RS is a diameter of the circle. NM \parallel RS  and \angle MRS= 29^o . Find (i) \angle RNM (ii) \angle NRM

Answer:

(i) \angle RMS = 90^o (angle in a semi circle)

\therefore \angle RMS = 180^o-90^o-29^o = 61^o

\therefore \angle RNM + 61^o = 180^o \Rightarrow \angle RNM = 119^o

c411.jpg(ii) Given NM \parallel RS

\angle NMR = \angle MRS = 29^o (alternate angles)

\angle NMS  = 90^o+29^o = 119^o

\angle NRS + \angle NMS = 180^o

\angle NRM+\angle MRS + \angle NMR + \angle RMS = 180^o

\angle NRM +29^o+29^o+90^o = 180^o

\Rightarrow \angle NRM = 180^o-148^o = 32^o

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c42Question 28: In the given figure, AB \parallel CD and O is the center of the circle. If \angle ADC = 25^o , find the  \angle AEB . Give reasons.

Answer:

\angle CAD = 90^o and \angle CBD = 90^o (angles in a semi circle)

Since AB \parallel CD

c49\angle CDA = \angle DAB = 25^o

\angle BAC = \angle BAD + \angle CAD

= 25^o+90^o=115^o

\therefore \angle ADB = 180^o-25^o -\angle ABD

In \triangle \angle ACD +90^o+25^o = 180^o

\Rightarrow \angle ACD = 180^o-115^o = 65^o

\angle ACD + \angle ABD = 180^o (cyclic quadrilateral)

\angle ABD = 180^o-65^o=115^o

\angle ADB = 180^o-25^o-115^o=40^o

\therefore \angle ADB = \angle AEB = 40^o (angles in the same segment)

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c41Question 29: Two circles intersect at P and Q . Through P a straight line APB is drawn to meet the circles in A and B . Through Q , a straight line is drawn to meet the circles at C and D . Prove that AC is parallel to BD .

Answer:

APQC and BPQD are cyclic quadrilateral

c410\angle CAP + \angle PQC = 180^o … … … … (i)

\angle PQD + \angle PBD = 180^o … … … … (ii)

\angle PQC + \angle PQD = 180^o   … … … … (iii)

From (i) and (ii)

\angle CAP + 180^o - \angle PQD = 180^o

\angle CAP = \angle PQD   … … … … (iv)

From (iv) and (iii)

\angle CAP = 180^o -\angle PBD

\Rightarrow \angle CAP + \angle PBD = 180^o     … … … … (v)

Therefore AB \parallel BD by (v)

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Question 30: ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD . Prove that AD is parallel to BC .

c45Answer:

ABCD is a cyclic quadrilateral, AD \parallel BC and PA = PD (given)

\angle PAD = \angle PDA

\angle CDA = 180^o - \angle PDA

\angle ABC + \angle CDA = 180^o

\angle ABC + 180^o - \angle PAD = 180

\angle ABC = \angle PAD

\therefore AD \parallel BC

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c591Question 31: AB is a diameter of a circle, APBR as shown in the figure. APQ and RBQ are straight lines. Find (i) \angle PRB (ii) \angle PBR (iii) \angle BPR

Answer:

(i) \angle BAP = \angle BRP = 35^o (angles in the same segment of the circle subtended by the same chord)

(ii) \angle ABQ = 180^o-35^o-25^o=120^o

\angle APB = 90^o (angle in a semi circle subtended by the diameter)

\angle PBA = 180^o-35^o-90^o = 55^o

\therefore \angle PBR = \angle PBA + \angle ABR = 55^o+60^o = 115^o

(iii) \angle BPR = 180^o-35^o-115^o = 30^o

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c59Question 32: In the given figure, SP is bisector of \angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ=SR .

Answer:

PQRS is a cyclic quadrilateral

\angle SPR = \angle SPT = x

\angle QPR = 180^o-2x

\angle SQR + \angle QPS = 180 (cyclic quadrilateral)

\angle SRQ = 180^o-(180^o-x)= x

Also \angle RQS = \angle RPS = x (angles in the same segment of the circle subtended by the same chord)

\therefore \angle RQS = \angle QRS

\therefore RS=QS

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c58Question 33: In the figure, O is the center of the circle, \angle AOE=150^o, \angle DAO=51^o . Calculate \angle CEB and \angle OCE .

Answer:

\angle ADE = \frac{1}{2} Reflex (\angle AOE) = \frac{1}{2} (360^o-150^o) = 105^o

\angle DAB + \angle DEB = 180^o

\angle DEB = 180^o-51^o=129^o

\angle CEB=180^o-129^o=51^o

\angle OCE = 180^o-51^o-105^o=24^o

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c57Question 34: In the given figure, P and Q are the centers of two intersecting circles intersecting at B and C . ACD is a straight line. Calculate the numerical value of x .

Answer:

\angle ACB = \frac{1}{2} \angle APB =75^o

ACD is a straight line

\angle BCD = 180^o-75^o = 105^o

\angle BCD = \frac{1}{2} (360^o-x)

\Rightarrow x = 360^o-210^o = 150^o

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c56Question 35: In the figure given below, two circles intersect at A and B . The center of the smaller circle is O and lines on the circumference of the larger circle. Given \angle APB = a . Find in terms of a the value of (i) Obtuse \angle AOB (ii) \angle ACB (iii) \angle ADB . Give reasons.

Answer:

(i) \angle APB = \frac{1}{2} (\angle AOB)

\angle AOB = 2a

(ii) \angle BOA + \angle ACB = 180^o \  (AOBC is a cyclic quadrilateral)

\therefore \angle ACB = 180^o-2a

(iii) \angle ADB = \angle ACB (angles in the same segment of the circle subtended by the same chord)

\therefore \angle ADB = 180^o-2a

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c55Question 36: In the given figure O is the cent of the circle and \angle ABC=55^o . Calculate x and y .

Answer:

\angle AOC = 2 \angle ABC \Rightarrow \angle AOC = 110^o = x

ABCD is a cyclic quadrilateral

\angle ADC + \angle ABC = 180^o

\Rightarrow y = 180^o-55^o=125^o

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c54Question 37: In the given figure, A is the center of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that \angle BCD = 2 \angle ABE .

Answer:

 \angle BAD = 2 \angle BED (angle subtended at the center is twice subtended on the circumference by the same chord)

 \angle BED = \angle ABE (alternate angles)

\therefore \angle BAD = 2 \angle ABE

ABCD is a parallelogram

\therefore \angle BAD = \angle BCD (opposite angles in a parallelogram are equal)

\angle BCD = 2 \angle ABE

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c53Question 38: ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given \angle BED = 65^o ; calculate: (i) \angle DAB (ii) \angle BDC .

Answer:

\angle BED = 65^o

\angle DAB = \angle DEB = 65^o (angles in the same segment of the circle subtended by the same chord)

\angle ADB = 90^o (angle subtended by the diameter on a semi circle)

\angle DBA = 180^o-65^o-90^o = 25^o

\therefore \angle BDC = 25^o (alternate angles)

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c52Question 39: In the given figure AB is the diameter of the circle.  Chord ED \parallel AB and \angle EAB=63^o . Calculate (i) \angle EBA (ii) \angle BCD .

Answer:

(i) \angle AEB = 90^o (angle in the semi circle)

\angle EBA = 180^o-90^o-63^o = 27^o

(ii) ED \parallel AB

\angle DEB = \angle EBA = 27^o (alternate angles)

EBCD is a cyclic quadrilateral

\angle DEB + \angle DCB = 180^o

\angle DCB = 180^o-27^o=153^o

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c51.jpgQuestion 40: The sides AB and CD of a cyclic quadrilateral ABCD are produced to meet at E , the sides DA and CB are produced to meet at F . If  \angle BEC = 42^o and \angle BAD=98^o find (i) \angle AFB (ii) \angle ADC

Answer:

\angle BAF = 180^o-98^o=82^o

\angle DCB = 180^o-98^o = 82^o

\angle BCE = 180^o-82^o = 98^o

\angle CBE = 180^o-98^o-42^o = 40^o

\angle ABF = 40^o (vertically opposite angles)

\angle AFB = 180^o-82^o-40^o=58^o

\angle CBA = 180^o-40^o = 140^o

\angle ADC = 180^o-140^o=40^o (cyclic quadrilateral)

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c61Question 41: In the given figure, AB is the diameter of the circle with center O . DO \parallel CB and \angle DCB = 120^o . Calculate (i) \angle DAB (ii) \angle DBA (iii) \angle DBC (iv) \angle ADC . Show that \triangle AOD is an equilateral triangle.

Answer:

(i) ABCD is a cyclic quadrilateral

\therefore \angle DAB + \angle DCB = 180^o

\Rightarrow \angle DAB = 60^o

(ii) \angle ADB = 90^o (angle in the  semicircle)

In \triangle ADO ,

\angle ADO = \angle AOD = 60^o

\therefore \angle ODB = 30^o

In \triangle \angle ODB, OD = OB 

\Rightarrow \angle DBA = 30^o

(iii) \angle ODB = 30^o

\therefore \angle DBC = 30^o (alternate angles)

(iv) \angle ABC = 60^o

\angle ADC + \angle ABC = 180^o (cyclic quadrilateral)

\angle ADC = 180^o-60^o = 120^o

In \triangle AOD ,

OD = OA (Radius of the same circle)

\therefore \angle ADO = \angle AOD = 60^o

Therefore all angles are 60^o . Hence \triangle AOD is an equilateral triangle.

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c62Question 42: In the given figure I is the incenter of the \triangle ABC . BI when produced meets the circumcenter of the the \triangle ABC at D . Given \angle BAC = 55^o and \angle ACB = 65^o . Calculate: (i) \angle DCA (ii) \angle DAC (iii) \angle DCI (iv) \angle AIC

Answer:

(i) IB is bisector of  \angle ABC

\angle ABD = \frac{1}{2} \angle ABC

\angle ABC = 180^o-55^o-65^o=60^o

\therefore \angle ABD = 30^o

(ii) \angle DAC = \angle CBD = 30^o (angles in the same segment)

(iii) \angle ACI = \frac{1}{2} \angle ACB

CI \ bisects \  \angle ACB

\therefore \angle ACI = \frac{65}{2} = 32.5^o

(iv) \angle IAC = \frac{1}{2} \angle BAC  = \frac{55}{2} = 27.5^o

AI bisects \angle BAC

\therefore \angle AIC = 180^o - \angle IAC- \angle ICA = 180^o-27.5^o-32.5^o = 120^o

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c63Question 43: A \triangle ABC is inscribed in a circle. The bisector of \angle BAC, \angle ABC and \angle ACB meet the circumference of the triangle at points P, Q and R respectively. Prove that (i) \angle ABC = 2 \angle APQ (ii) \angle ACB = 2 \angle APR (iii) \angle QPR = 90^o- \frac{1}{2} \angle BAC

Answer:

(i) BQ bisects \angle ABC

\therefore \angle ABQ = \frac{1}{2}  \angle ABC

\angle APQ = \angle ABQ (angle in same segment)

\therefore \angle ABC = 2 \angle ABQ = 2 \angle APQ  … … … … (i)

(ii) CR bisects \angle ACB

\therefore \angle ACR = \frac{1}{2} \angle ABC

and \angle ACR = \angle APR (angles in the same segment)

\therefore \angle ACB = 2 \angle APR … … … … (ii)

(iii) Adding (i) and (ii)  we get

\angle ABC + \angle ACB = 2 (\angle APQ + \angle APR) = 2 \angle QPR

\Rightarrow 180^o - \angle BAC = 2 \angle QPR

\Rightarrow \angle QPR = 90^o-\frac{1}{2} \angle BAC

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c64Question 44: Calculate angles x, y, z if: \frac{x}{3}= \frac{y}{4}=\frac{z}{5}

Answer:

\frac{x}{3}= \frac{y}{4}=\frac{z}{5} = k \ (say)

\Rightarrow x = 3k, y = 4k \ and \ z = 5k

\angle ABC + \angle ADC = 180^o 

\angle PBC = 180^o-3k-4k = 180^o-7k

\therefore ABC = 7k

Similarly \angle ADC = 180^o -(180^o-3k-5k) = 8k

\therefore \angle ABC + \angle ADC = 7k+8k = 15k

15 = 180^o \Rightarrow k = 12

Therefore x = 36^o, y = 48^o \ and \ z = 60^o

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c65Question 45: In the given figure AB = AC = CD and \angle ADC = 38^o , calculate (i) \angle ABC (ii) \angle BEC [1995]

Answer:

(i) AC = CD

\angle CAD = \angle CDA = 38^o

\angle ACD = 180^o-38^o - 38^o = 104^o

\angle ACB = 180^o-104^o = 76^o

AB = AC

\angle ABC = \angle ACB = 76^o

(ii) \angle BAC = 180^o-76^o-76^o = 38^o

\angle BAC = \angle BEC = 38^o (angles in the same segment)

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c66Question 46: In the given figure AC is the diameter of the circle with center O . Chord BD \perp AC . Write down the angles p, q and r in terms of x . [1996]

Answer:

\angle AOB = 2 \angle ACB = 2 \angle ADB (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

x = 2q \Rightarrow q=\frac{x}{2}

\angle ADB = \frac{x}{2}

\angle ADC = 90^o (angles in a semicircle)

r +\frac{x}{2} = 90^o \Rightarrow r = 90-\frac{x}{2}

Now \angle DAC = \angle DBC (angles in the same segment)

p = 90^o-q \Rightarrow p = 90^o-\frac{x}{2}

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c67Question 47:  In the given figure AC is the diameter of the circle with center O . $CD \parallel BE &s=0$. \angle AOB = 80^o and \angle ACE = 110^o . Calculate (i) \angle BEC (ii) \angle BCD (iii) \angle CED [1998]

Answer:

(i) \angle BOC = 180^o-80^o=100^o (Since AC is a straight line)

\angle BOC = 2 \angle BEC   (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\Rightarrow \angle BEC = \frac{100}{2} = 50^o

(ii) DC \parallel EB

\angle DCE = \angle BEC = 50^o

\angle AOB = 80^o (given)

\Rightarrow \angle ACB = \frac{1}{2} \angle AOB = 40^o  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\angle BCD = \angle ACB + \angle ACE + \angle DCE = 40^o+10^o+50^o = 100^o

(iii) \angle BED = 180^o-\angle BCD = 180^o-100^o = 80^o (cyclic quadrilateral)

\angle CED + 50^o = 80^o \Rightarrow \angle CED = 30^o

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c68Question 48: In the given figure, AE is the diameter of the circle. Write down the numerical value of \angle ABC + \angle CDE . Give reasons for your answer. [1998]

Answer:

\angle AOC = \frac{180}{2} = 90^o

\angle AOC = 2 \angle AEC  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\Rightarrow \angle AEC = \frac{90}{2} = 45^o

ABCE is a cyclic quadrilateral

\therefore \angle ABC + \angle AEC = 180 ^o

\Rightarrow \angle ABC = 180-45 = 135^o

Similarly, \angle CDE = 135^o

\therefore \angle ABC + \angle CDE = 135^o + 135^o = 270^o

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c69.jpgQuestion 49: In the given figure AOC is the diameter and AC \parallel ED . If \angle CBE = 64^o , calculate \angle DEC . [1991]

Answer:

\angle ABC = 90^o (angle in a semi circle)

\angle ABE = 90^o-64^o = 26^o

\angle ABE = \angle ACE = 26^o (angles in the same segment)

AC \parallel ED

\therefore \angle DEC = \angle ACE = 26^o (alternate angles)

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c610Question 50: Use the given figure below to find (i) \angle BAD (ii) \angle DQB  [1987]

Answer:

(i) In \triangle ADP 

\angle BAD = 180^o - 85^o - 40^o = 55^o 

(ii) \angle ABC = 180^o - \angle ADC = 180^o - 85^o = 95^o 

\angle AQB = 180^o - 95^o - 55^o = 30^o 

\Rightarrow \angle DQB = 30^o 

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c79Question 51: In the given figure AOB is the diameter and DC \parallel AB . If \angle CAB = x^o , find in terms of x , the values of: (i) \angle COB (ii) \angle DOC (iii) \angle DAC (iv) \angle ADC [1991]

Answer:

(i) \angle OCB = 2 \angle CAB = 2x (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

(ii) \angle OCD = \angle COB = 2 x (alternate angles)

In \triangle OCD 

OC = OC (radius of the same circle)

\angle ODC = \angle OCD = 2x

\angle DOC = 180^o-2x-2x = 180^o-4x

(iii) \angle DAC = \frac{1}{2} \angle DOC = \frac{1}{2} (180^o-4x) = 90^o-2x  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

(iv) DC \parallel AO (given)

\therefore \angle ACD = \angle OAC = x (alternate angles)

\therefore \angle ADC = 180^o - \angle DAC-\angle ACD = 180^o-(90^o-2x) - x = 90^o+x

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c78Question 52: In the figure AB is the diameter of the circle with center O . \angle BCD = 130^o .  Find (i) \angle DAB (ii) \angle DBA [2012]

Answer:

(i) \angle DAB = 180^o-\angle DCB = 180^o-130^o = 50^o  \ (ABCD  is a cyclic quadrilateral)

(ii) In \triangle ADB 

\angle DAB + \angle ADB + \angle DBA = 180^o 

\Rightarrow \angle DBA = 180^o-50^o-90^o = 140^o 

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c77Question 53: In the given figure PQ is the diameter of the circle whose center is O . Given \angle ROS=42^o , calculate \angle RTS . [1992]

Answer:

Join P and S as shown in the diagram.

\angle PSQ = 90^o (angle in a semi circle)

c715\angle SPR = \frac{1}{2} \angle ROS

\Rightarrow \angle SPT = \frac{1}{2} \times 42^o = 21^o

In \triangle PST

\angle PST = 90^o-\angle SPT

\Rightarrow \angle RTS = 90^o-21^o = 69^o

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c76Question 54: In the given figure PQ is the diameter. Chord SR \parallel PQ . Given the \angle PQR = 58^o , calculate (i) \angle RPQ (ii) \angle STP [1989]

Answer:

c714Join P and R as shown in the diagram.

(i) \angle PRQ = 90^o (angles in the semi circle)

\therefore \angle RPQ = 90^o- \angle PQR = 90^o-58^o = 32^o

(ii) SP \parallel PQ (given)

\angle PSR = \angle RPQ = 32^o (alternate angles)

\angle SPT = 180^o-\angle PSR = 180^o-32^o = 148^o (PTSR is a cyclic quadrilateral)

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c75Question 55: AB   is the diameter of the circle with center O  . OD \parallel BC   and \angle AOD = 60^o  . Calculate the numerical values of (i) \angle ABD   (ii) \angle DBC   (iii) \angle ADC [1987]

Answer:

Join B and D as shown in the diagram.

(i) \angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 60^o = 30^o  (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)c713

(ii) \angle BDA = 90^o (angle in semi circle)

Since \angle OAD = 60^o \Rightarrow \triangle OAD  is equilateral

\angle ODB = 90^o-ODA = 90^o-60^o = 30^o

Since OD \parallel BC

\angle DBC = \angle ODB = 30^o (alternate angles)

(ii) \angle ABC = \angle ABD + \angle DBC = 30^o +30^o = 60^o

\angle ADC = 180^o -\angle ABC = 180^o- 60^o = 120^o ( ABCD is a cyclic quadrilateral)

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c74Question 56: In the given figure, the center O of the smaller circle lies on the circumference of the bigger circle. If \angle APB = 75^o and  \angle BCD = 40^o , find: (i) \angle AOB (ii) \angle ACB (iii) \angle ABD (iv) \angle ADB . [1984]

Answer:

Join A and B as shown in the diagram.

(i) \angle AOB = 2 \angle APB = 2 \times 75^o = 150^o  (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)c712

(ii) \angle ACB = 180^o-\angle AOB^o = 180^o-150^o = 30^o  (AOBC  is a cyclic quadrilateral)

(iii) \angle ABD = 180^o - \angle ACD = 180^o-(40^o+30^o) = 110^o  (ABDC  is a cyclic quadrilateral)

(iv) \angle ADB = 180^o - \angle AOB = 180^o-150^o=  30^o  (ADBO is a cyclic quadrilateral)

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c73Question 57: In the figure, \angle BAD = 65^o , \angle ABD = 70^o and \angle BDC=45^o . Find (i) \angle BCD (ii) \angle ACB . Hence show that AC is the diameter.

Answer:

(i) \angle BCD = 180^o-\angle BAD = 180^o-65^o=115^o (ABCD is a cyclic quadrilateral)

(ii) In \triangle ABD

\angle ADB = 180^o-65^o-70^o=45^o

\angle ACB = \angle ADB = 45^o (angles in the same segment)

\angle ADC = \angle ADB + \angle BDC = 45^o+45^o = 90^o

Therefore AC is the diameter since the angle subtended by AC on the circumference is 90^o .

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Question 58:  In a cyclic quadrilateral ABCD , \angle A : \angle C = 3:1 and \angle B : \angle D = 1:5 . Find each angle of the quadrilateral.

Answer:

\angle A : \angle C = 3:1 \Rightarrow \angle A = 3x  \ and \ \angle C = x

\angle A + \angle C = 180^o (ABCD is a cyclic quadrilateral)

\therefore 3x + x = 180^o \Rightarrow x = 45^o

\therefore \angle A = 135^o \ and \ \angle C = 45^o

Also given \angle B:\angle D = 1:5 \Rightarrow \angle B = y \ and \ \angle D = 5y 

\angle B + \angle D = 180^o (ABCD is a cyclic quadrilateral)

\therefore y + 5y = 180^o \Rightarrow y = 30^o

\therefore \angle B = 30^o \ and \ \angle D = 150^o

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c72Question 59: The given figure shows a circle with center O and \angle ABP=42^o . Find (i) \angle PQB (ii) \angle QPB + \angle PBQ

Answer:

Join A and P as shown in the diagram.

(i) \angle APB = 90^o (angle in semi circle)

c711\therefore \angle BAP = 90^o-\angle ABP = 90^o-42^o = 48^o

Now, \angle PQB = \angle BAP = 48^o (angles in the same segment)

(ii) in \triangle BPQ

\angle QPB + \angle PBQ = 180^o-\angle PQB = 180^o-48^o = 132^o

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c71Question 60: In the given figure M is the center of the circle. Chords AB and CD are perpendicular to each other. If \angle MAD = x and \angle BAC = y (i) express \angle AMD in terms of x   (ii) express \angle ABD in terms of y (iii) prove that x = y

Answer:

Given: AB \perp CD

\angle MAD = x \ and \  \angle BAC = y

(i) In \triangle AMD

MA = MD

\angle MAD = \angle MDA = x

In \triangle AMD

\angle MAD + \angle MDA + \angle AMD = 180^o

\Rightarrow x + x + \angle AMD = 180^o

\angle AMD = 180^o - 2x

(ii) \angle AMD = 2 \angle ABD (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)

\angle ABD = \frac{1}{2} (80^o-2x)

\Rightarrow \angle ABD = 90^o-x

AB \perp CD, ALC = 90^o

in \triangle ALC

\angle LAC + \angle LCA = 90^o

\Rightarrow \angle BAC + \angle DAC = 90^o

\Rightarrow \angle DAC = 90^o-y

Now, \angle DAC = \angle ABD (angles in the same segment)

\angle ABD = 90^o-y

(iii) \angle ABD = 90^o-y and we proved that \angle ABD = 90^o-x

\therefore 90^o-x = 90^o-y \Rightarrow x = y

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Question 61: In a circle, with center O , a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P , show that \angle APB = 60^o .

c710Answer:

In \triangle OCD, OD = OC = DC

\Rightarrow \triangle OCD is equilateral

\therefore \angle ODC = 60^o

\angle ADC + \angle ABC = 180^o (ABCD is a cyclic quadrilateral)

\Rightarrow \angle ODA + 60^o + \angle ABP = 180^o

\Rightarrow \angle OAD + \angle ABP = 90^o

\Rightarrow \angle PAB + \angle ABP = 120^o

In \triangle PAB 

\angle APB = 180^o - \angle PAB - \angle ABP = 180^o-120^o = 60^o