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

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

$\Rightarrow AB = BE$

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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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