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 $\angle OCE$.

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

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

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

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

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

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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 $x$.

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

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

$\angle AOB = 2a$

(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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