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

In $\triangle AOC$

$\angle OAC = \angle OCA = x^o$

Therefore $\angle AOC + 2x^o = 180^o$

$\Rightarrow \angle AOC = 180^o-2x^o$

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)

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

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

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Question 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]

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

$OA = OC =$  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) (ii)

(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) (ii) (iii) (iv)

(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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Question 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.

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

$ABCD$ 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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Question 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) $\angle CBA$

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

Since $OA = OB$ (radius of the same circle)

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

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

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

$\angle CDB = \angle BAC = 49^o$ (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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Question 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)  $\angle BCA$

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

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

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

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

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

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

$\angle ACB = \angle ADB$ (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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Question 12: Given: $\angle CAB = 75^o$ and $\angle CBA=50^o$. Find the value of $\angle DAB + \angle ABD$.

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

$\angle ADB = 55^o$ (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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Question 13: $ABCD$ is a cyclic quadrilateral in circle with center $O$. If $\angle ADC = 130^o$; find $\angle BAC$.

$\angle ADC = 130^o$

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

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

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

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

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

$OA = OC$ (radius of the same circle)

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

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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Question 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) $\angle CAB$

$\angle ADB = \angle ACB = 30^o$

(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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Question 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 $\angle COF = \angle CEF$.

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

$\angle CEF = 64^o$

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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Question 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) $\angle ABC$

$\angle CED = 90^o$ (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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Question 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 $AB = BE$.

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

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

$\therefore \triangle ABO \sim \triangle AEC$

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

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

$\Rightarrow AB = BE$

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

(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.

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

$AB = AC$ (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.

$\angle PQA = 90^o$ (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: $\angle OBD+\angle ODB= \angle CBD+\angle CDB$.

$\angle BAD + \angle BCD = 180^o$ (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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Question 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.

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

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

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

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

(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 $\angle APC$.

$\angle AOC = 2 \angle ADC$

Similarly, $\angle BOD = 2 \angle BAD$

$\angle AOC + \angle BOD = 2(\angle ADC+\angle BAD)$ … … … … (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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Question 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) $\angle NRM$

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

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

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

(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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Question 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.

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

Since $AB \parallel CD$

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

$\angle BAC = \angle BAD + \angle CAD$

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

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

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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Question 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 $BD$.

$APQC$ and $BPQD$ are cyclic quadrilateral

$\angle CAP + \angle PQC = 180^o$ … … … … (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 $BC$.

$ABCD$ 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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Question 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) $\angle BPR$

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

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

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

$\angle DAB + \angle DEB = 180^o$

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

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

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

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Question 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 $x$.

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

$ACD$ is a straight line

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

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

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

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Question 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.

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

$\angle AOB = 2a$

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

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

$ABCD$ is a cyclic quadrilateral

$\angle ADC + \angle ABC = 180^o$

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

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

$\angle BAD = 2 \angle BED$ (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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Question 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) $\angle BDC$.

$\angle BED = 65^o$

$\angle DAB = \angle DEB = 65^o$ (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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Question 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) $\angle BCD$.

(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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Question 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) $\angle ADC$

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

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

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

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

$\angle ABF = 40^o$ (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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Question 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.

(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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Question 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) $\angle AIC$

(i) $IB \text{ is bisector of } ABC$

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

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

$\therefore \angle ABD = 30^o$

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

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

$CI \text{ bisects } \angle ACB$

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

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

$AI$ 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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Question 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) $\displaystyle \angle QPR = 90^o- \frac{1}{2} \angle BAC$

(i) $BQ$ bisects $\angle ABC$

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

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

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

(ii) $CR$ bisects $\angle ACB$

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

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

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

(iii) Adding (i) and (ii)  we get

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

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

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

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

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

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

$\angle ABC + \angle ADC = 180^o$

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

$\therefore ABC = 7k$

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

(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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Question 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]

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

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

$\displaystyle \angle ADB = \frac{x}{2}$

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

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

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

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

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Question 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]

(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)

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

(ii) $DC \parallel EB$

$\angle DCE = \angle BEC = 50^o$

$\angle AOB = 80^o$ (given)

$\displaystyle \Rightarrow \angle ACB = \frac{1}{2} \angle AOB = 40^o$ (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)

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

$\\$

Question 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]

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

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

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

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

$\angle ABC = 90^o$ (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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Question 50: Use the given figure below to find (i) $\angle BAD$ (ii) $\angle DQB$ [1987]

(i) In $\triangle ADP$

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

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

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

$\Rightarrow \angle DQB = 30^o$

$\\$

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) $\displaystyle \angle DAC = \frac{1}{2} \angle DOC = \frac{1}{2} (180^o-4x) = 90^o-2x$ (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)

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

$\\$

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 $\triangle ADB$

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

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

$\\$

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)

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

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

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) $\displaystyle \angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 60^o = 30^o$ (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)

$\\$

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)

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

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

$\\$

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 $\triangle BPQ$

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

$\\$

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)

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

$\Rightarrow \angle ABD = 90^o-x$

$AB \perp CD, ALC = 90^o$

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 $\angle ABD = 90^o-x$

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

$\\$

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 $\triangle PAB$

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