Question 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]

(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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Question 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]

(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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Question 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]

Join P and S as shown in the diagram.

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

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

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

In $\triangle PST$

$\angle PST = 90^o-\angle SPT$

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

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Question 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]

Join 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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Question 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]

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)

(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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Question 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]

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)

(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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Question 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.

(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.

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

$\angle A + \angle C = 180^o$ ($ABCD$ 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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Question 59: The given figure shows a circle with center $O$ and $\angle ABP=42^o$. Find (i) $\angle PQB$ (ii) $\angle QPB + \angle PBQ$

Join A and P as shown in the diagram.

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

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

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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Question 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 $x = y$

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 $\angle APB = 60^o$.

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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