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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 13 Angle Properties of a Circle
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 13, Angle Properties of a Circle. Here students can easily find step by step solutions of all the problems for Angle Properties of a Circle, Exercise 13.1, 13.2, 13.3 and 13.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Angle Properties of a Circle Exercise 13.1 Solution
Question no – (1)
Solution :
(i) m AC = 65°
(ii) m BC = 180° – 65° = 115°
(iii) m AB = 180°
(iv) m ABC = 360° – 65° = 295°
Question no – (2)
Solution :
Let the minor are subtend an angle of x at the centre. Then the major are will subtend an angle of 3x°
Now, 3x° + x° = 360°
Or, 4x = 360°
Or, x = 90°
So, the minor are will subtend an angle of 90° and the major are will subtend an angle of,
= 360° – 90°
= 270°
Question no – (3)
Solution :
Let the angle subtended by the minor are be x°
Now, 2 × m (major arc) = 4 × m (minor arc)
Or, m (major arc) = 2 × m (minor arc)
∴ The angle subtended by the major are will be 2x°
∴ 2x° + x° = 360°
Or, 3x° = 360°
Or, x° = 360°/3
x° = 120°
and 2x° = 120° × 2
= 240°
Question no – (5)
Solution :
Yes, m (minor AB) = m (minor CD) because, the length of arc AB and arc CD are equal.
Angle Properties of a Circle Exercise 13.2 Solution
Question no – (1)
Solution :
Required Figure :
(i) ∠y = 1/2 ∠x
= 1/2 × 120°
= 60°
(ii) ∠x = 2 ∠y
= 2 × 85°
= 170°
Question no – (2)
Solution :
Considering ∆AOB, ∠OAB + ∠OBA = 180° – 80° = 100°
Or, 2 (∠OAB) = 100°
Or, ∠OAB = 1000/2 = 50°
Considering ∆AOC, ∠AOC = 180° – 80° = 100°
∴ ∠OAC + ∠OAC = 180° – 100° = 80°
Or, 2 (∠OAC) = 80°
Or, ∠OAC = 80°/2 = 40°
Question no – (3)
Solution :
Required Figure :
Since, ∆ABC is an equilateral triangle,
∠BAC = 60°
Now, ∠BOC = 2 × 60°
∠BOC = 120°
Question no – (4)
Solution :
Required Figure :
Since, ∠BAC is an angle in a semicircle, ∠BAC = 90°
∴ ∠ACB = 180° – (90° + 35°)
= 180° – 125°
= 55°
∴ ∠ACB = 55°
Question no – (5)
Solution :
Required Figure :
(i) ∠ACD = ∠CAB
(ii) m AD = ∠ACD, m BC = ∠CAB
Therefore, m AD2 ≅ m BC
Question no – (6)
Solution :
∠AOB = 2 × ∠ABC
= 2 × 45°
= 90°
∴ ∠AOB = 90°
Question no – (7)
Solution :
Required Figure :
m AC = ∠AOC = 150,
m AB = ∠AOB = 120°,
∴ m CB = ∠COB
= 360° – (150° + 120°)
= 360° – 270°
= 90°
∴ ∠CAB = 1/2 ∠COB = 1/2 × 90° = 45°
∴ ∠ABC = 1/2 ∠AOC = 1/2 × 150° = 75°
∴ ∠ACB = 1/2 ∠AOB = 1/2 × 120° = 60°
Question no – (8)
Solution :
∠AOB = 110°,
∴ ∠BOC = 180° – 110°
= 70°
Now, are CB subtends ∠BDC at any other point of the circumference
∠BDC = 1/2 × ∠BOC
= 1/2 × 70°
= 35°
Question no – (9)
Solution :
M (AM) = ∠ACM = 1/2 ∠AOM,
m (MB) = ∠MCB = 1/2 ∠MOB
Now, Since ∠ACM = ∠MCB,
Therefore, m (AM) = m (MB)
Question no – (10)
Solution :
Considering ∆BOC, OC = CB = OB [OC,OB, are radius and CB = OC]
Since all the three sides are equal to each other,
∠COB = 60°
Now, ∠BAC = 1/2 ∠BOC
= 1/2 × 60°
= 30°
∴ ∠BAC = 30°
Question no – (11)
Solution :
Since, AB = AC, ∆ABC is an isosceles triangle, in which ∠BAC = 90°
∴ ∠ACB + ∠ABC = 90° and ∠ACD
= ∠ABC, Since AB = AC
∴ ∠ACB = ∠ABC
= 90°/2
= 45°
∴ ∠ACB = 45°
Question no – (12)
Solution :
Since, ∠ACB, ∠ADC are angles in hemisphere both are 90°
∴ ∠ABC = 90° – 45° = 45°
∴ ∠CBA = ∠ABC + ∠ABD
= 45° + 55°
= 100°
Now, ∠BAD = 90° – 55° = 35°
∴ ∠CAD = ∠BAD + ∠BAC
= 45° + 35°
= 80°
Question no – (13)
Solution :
m (DB) ∠BOD = 30°, m (BC) = ∠BOC = 80°
Since, ∠BOC and ∠BAC are on, Same are, but ∠BOC is at centre and ∠BAC is at other point, ∠BOC = 2 ∠BAC
∴ ∠BAC = 1/2 × 80° = 40°
Similarly, ∠BAD = 1/2 ∠BOD = 1/2 × 30° = 15°
∠AOC = 180° – (∠BOD + ∠BOC) = 180° – (30° + 80°)
= 180° – 110° = 70°
Question no – (14)
Solution :
Since, AO = OB = OC ∠OAB = ∠OBA = 30° and ∠OBC = ∠OCB = 40°
∴ ∠ABC = 30° + 40° = 70°
∴ ∠AOC = 2 ∠ABC = 2 × 70° = 140°
Question no – (15)
Solution :
In ∆BOC, ∠BOC + ∠OAC = 360° – (90° + 110°)
= 360° – 200°
= 160°
Now, since OB = OC, ∠OBC = ∠OCB,
∠OBC = 180° – 160°/2
= 20°/2
= 10°
Question no – (16)
Solution :
Since, OC = OA, ∠OCA
= ∠OAC and ∠OCA + ∠OAC
= 180° – 110°
= 70°
∴ ∠OCA = ∠OAC = 70°/2 = 35°
Now, m ∆ABO, OA = OB,
Therefore ∠OAB = ∠OBA and ∠OAB + ∠OBA = 90°
∴ ∠OAB = ∠OAB = 90°/2 = 45°
∴ ∠BAC = ∠BAO + OAC
= 45° + 35°
= 80°
Angle Properties of a Circle Exercise 13.3 Solution :
Question no – (1)
Solution :
(i) x = 35
Question no – (2)
Solution :
Since ∠ABC and ∠ADC are on same arc,
∠ABC = ∠ADC = 35°
Question no – (3)
Solution :
Required figure,
∠BAD = ∠BCD = 25°
Now, in ∆BCE, ∠CBE = 180° – (25° + 30°) = 180° – 55°
= 125°
Question no – (4)
Solution :
Required figure,
∠ABC = ∠AEC = 30°
Now, m ∆CDE, ∠CDE = 90° and ∠DEC = 30°
∴ ∠DCE = ∠BCE = 180° – (90° – 130°) = 180° – 120° = 60°
Question no – (5)
Solution :
Required figure,
∠CAB = ∠CDB = 35°
Since, ∠CAB and ∠CDB are on same arc.
∴ ∠CDB = 35°
Now, ∠ABD = 180° – (100 + 35°) = 180° – 135° = 45°
Question no – (6)
Solution :
Required figure,
∠CPB = ∠CAB = 35°
∴ ∠ABC = 180° – (115° + 35°)
= 180° – 150° = 30°
Question no – (7)
Solution :
Required figure,
∠CBE = ∠CAD = 30°
Since BD is bisector of ∠ABC, ∠ABD = 30° and thus ∠ABC = ∠ABD + ∠CBE = 30° + 30° = 60°
Question no – (8)
Solution :
Required figure,
CB seems parallel AD. So, (i) ∠APB = ∠BPC
(ii) ∠PAD = ∠PBC
(iii) ∠PDA = ∠PCB
So, I can say, ∆BPC and ∆APD are equiangular.
Question no – (9)
Solution :
Required figure,
∠ECB = ∠BAD = 25°
And ∠AED = 180° – ∠CEA
= 180° – 40° = 140°
∴ ∠ADE = 180° – (∠EAD + ∠AED)
= 180° – (25° + 140°) = 180° – 165° = 15°
∴ ∠ADE = ∠ADC = ∠ABC = 15°
Question no – (10)
Solution :
Required figure,
∠AOB = 2 ∠ACB = 2 × 60° = 120°
Now, ∠COD = ∠AOB
∴ ∠CBD = 120°/2 = 60°
Now, ∠ABO = 180° – 120°/2 = 60°/2 = 30°
∴ ∠ABD = ∠ABO + ∠OBD = 30° + 60° = 90°
Question no – (11)
Solution :
Required figure,
Since AL bisects ∠BAC, ∠OAC = 40°
Now, ∠ADC = 180° – (40 + 60°)
= 80°
Since, CK is the bisector of ∠ACO, ∠KCO = 30°
∴ ∠LKC = 180° – (30° + 80°) = 70°
Now, ∠LOC = 180° – 70° = 110°
∴ ∠OCL = 180° – 110°/2 = 35°
∴ ∠KCL = ∠KCO + ∠OCL = 30° + 35° = 65°
Question no – (12)
Solution :
Required figure,
∠BAC = ∠BDC = 30°
∴ ∠BCA = 180° – (∠BAC + ∠ABC)
=180° – (30° + 110°)
= 180° – 140° = 40°
Question no – (13)
Solution :
Required figure,
∠BAC = ∠BDC = 30°
∠CBD = ∠CAD = 70°
∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
Since, ∠BAD = 30° + 70° = 100°
∴ ∠BCD = 180° – 100° = 80°
Angle Properties of a Circle Exercise 13.4 Solution
Question no – (1)
Solution :
Required figure,
Since ∠A and ∠C are opposite angles in a cyclic quadrilateral ∠A + ∠C = 180°
∴ ∠C = 180° – 100° = 80°
∴ ∠BCE = 180° – ∠BCD
= 180° – 80°
= 100°
Question no – (2)
Solution :
Required figure,
Since, ∠B and ∠D are opposite angles in a cyclic quadrilateral,
∴ ∠B + ∠D = 180°
Or, ∠D = 180° – 70° = 110°
Similarly, ∠A + ∠C = 180°
Or, ∠C = 180° – ∠A
= 180° – 85°
= 95°
Question no – (3)
Solution :
Question no – (4)
Solution :
Since, ∠CAP and ∠CQP are opposite angles in cyclic quadrilateral, therefore, ∠x + ∠y = 180°
∴ ∠x = 180° – ∠CQP
= 180° – 95°
= 85°
Now, ∠x and ∠y are opposite angles in a cyclic quadrilateral, therefore ∠x + ∠y = 180°
∴ ∠y = 180° – ∠x
= 180° – 85°
= 95°
Now, ∠ACO and ∠APQ are opposite angles in a cyclic quadrilateral,
∴ ∠w = 180° – 70° = 110°
∴ Now, ∠QPB + ∠w = ∠QPB = ∠z = 180°
∴ ∠w = ∠z = 110°
Question no – (5)
Solution :
Required figure,
Since, ∠D and ∠B are opposite angles in a cyclic quadrilateral,
∠B = 180° – ∠D = 180° – 140° = 40° And, being an angle in a semicircles ∠ACB = 90°
∴ ∠BAC = 180° – (∠ACB + ∠ABC)
= 180° – (90° + 40°)
= 50°
Question no – (6)
Solution :
Required figure,
(i) ∠ADC = 180° – ∠ABC = 180° – 140°
(ii) ∠ADC = ∠CBE = 100°
(iii) ∠A = 180° – ∠C = 180° – 75° = 105°,
∠D = 180° – ∠B = 180° – 65° = 115°
(iv) We have, ∠ADC = 115°, ∠ACB = 90°
Now, ∠CBA = 180° – ∠ADC = 180° – 115° = 65°
∴ ∠BAC = 180° – (∠ACB + ∠ABC)
= 180° – (90° + 65°)
= 180° – 155°
= 25°
Question no – (7)
Solution :
Required figure,
We have ∠A + ∠C = 180° …….(i)
∠B + ∠D = 180° ……..(ii)
[Being opposite angles in a cyclic quadrilateral]
From (i), x + 2x = 180°
Or, 3x = 180°
Or, x = 180°/3 = 60°
From, (ii) 2y + 3y = 180°
Or, 5y = 180°
Or, y = 180°/5
∴ y = 36°
Question no – (8)
Solution :
Required figure,
Being opposite angles in a cyclic quadrilateral, we have,
Or, ∠AFE + ∠ABW = 180°
Or, ∠ABE = 180° – 95° = 85° and, ∠FAB + ∠FEB = 180°
Or, ∠FEB = 180° – 65° = 115°
Thus = ∠BED = 180° – ∠BEF = 180° – 115° = 65° = w
And, we also have x = y = 85°
And, w + z = 180°
Or, z = 180° – w =180° – 65° = 115°
Question no – (9)
Solution :
Required figure,
∠ABD = ∠ACD = 60° [angle on same arc]
In ∆CDP, ∆CPD = 90° = ∠APB [reflex angle]
∴ ∠CDP = 180° – (90° + 60°) = 30°
∠BDC = ∠BAC = 30° [angle on same arc]
We have ∠EDF = 85° = ∠CDA [reflex angle]
Now, ∠PDA = ∠CDA – ∠CDP = 85° – 30° = 55°
∠CAB = 30°
Now, ∠CDA + ∠CBA = 180° [Being opposite angles in a cyclic quadrilateral]
∴ ∠CBA = 180° – 85° = 95°
Now, ∠CBD + ∠DBA = ∠CBA = 95°
Or, ∠CBD = 95° – 60°
= 35°
Previous Chapter Solution :