c21Question 11: In the following figure, O is the center and \triangle ABC is equilateral. Find (i) \angle ADB (ii) \angle AEB

Answer:

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

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

c31\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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c22Question 12: Given: \angle CAB = 75^o and \angle CBA=50^o . Find the value of \angle DAB + \angle ABD .

Answer:

\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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c23Question 13: ABCD is a cyclic quadrilateral in circle with center O . If \angle ADC = 130^o ; find \angle BAC .

Answer:

\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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c24Question 14: In the given figure, AOB is a diameter of the circle and \angle AOC=110^o . Find \angle BDC .

Answer:

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

OA = OC (radius of the same circle)

c32Therefore 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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c25Question 15: In the following figure, O is the center of the circle, \angle AOB=60^o and \angle BDC=100^o . Find \angle OBC .

Answer:

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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c26Question 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

Answer:

\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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c27Question 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 .

Answer:

\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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c28Question 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

Answer:

\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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c29Question 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 .

Answer:

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

c33\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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c210Question 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 .

Answer:

(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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