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)

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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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) $\angle CBA$

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

Since $OA = OB$ (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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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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