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

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