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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