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Joy of Mathematics Class 6 Solutions Chapter 13 Angles
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Joy of Mathematics Class 6 Book, Chapter 13 Angles. Here students can easily find step by step solutions of all the problems for Angles. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Students will get chapter 13 solutions. Here students will find exercise wise solutions for chapter 13 Exercise 13.1 and 13.2
Angles Exercise 13.1 Solution :
Question no – (1)
Solution :
(a) Corner of walls.
(b) Pages of Book
Question no – (2)
Solution :
Vertex = Q
Arms = PQ, QR
Question no – (3)
Solution :
(a) Three angles
∠ABC, ∠BAC, ∠ACB
(b) 4 angles that’s are,
∠WXY, ∠XYZ, ∠YZW, ∠ZWX
(c) 8 angles have,
∠PQR, ∠QRS, ∠SPQ, ∠RSP, ∠PQS, ∠PSQ, ∠RSQ, ∠RQS
Question no – (4)
Solution :
(a) L, Q, P
(b) A, B, T, M
(c) S, X, P, Z
Question no – (5)
Solution :
(a) 10/60
= 1/6
(b) 20 minutes
= 20/60
= 1/3
(c) 40 minutes
= 40/60
= 2/3
Question no – (6)
Solution :
(a) 3 ‘o clock
= 90°
(b) 5 ‘o clock
= 5/12 × 360
= 150°
(c) 11 ‘o clock
= 11/12 × 360
= 330°
Question no – (7)
Solution :
(a) 1/5 turn
= (1/5 × 360)
= 72°
(b) 5/12 turn
= (5/12 × 360°)
= 150°
(c) 7/24 turn
= (7/24 × 360°)
= 105°
Question no – (8)
Solution :
(a) 75° = (75 × 60) = 4500’
= (4500 × 60)
= 270000”
(b) 17° = (17 × 60)’
= 1020’
= (1020 × 60)”
= 61200”
(c) 320° = (320 × 60)’
= 19200’
= (19200 × 60)”
= 1152000”
(d) 127° = (127 × 60)’
= 7620”
= (7620 × 60)”
= 457200”
Question no – (9)
Solution :
(a) 135°
(b) 180°
Question no – (10)
Solution :
(a) Reflex angle
(b) Obtuse angle
(c) Right angle
(d) Complete angle
Question no – (12)
Solution :
(a) 30°
(b) 45°
(c) 65°
(d) 90°
(e) 120°
Question no – (13)
Solution :
(a) Reflex angle
= (360° – 30°)
= 330°
(b) Reflex angle
= (360° – 140°)
= 220°
(c) Reflex angle
= (360° – 90°)
= 270°
(d) Reflex angle
= (360° – 150°)
= 210°
(e) Reflex angle
= (360° – 45°)
= 315°
(f) Reflex angle
= (360° – 135°)
= 225°
Question no – (14)
Solution :
2∠A + 4∠B
= 2(12°52’39”) + 4(15°24’33”)
= 2(12°) + 2(52’) + 2(39)” + 4(15°) + 4(24’) + 4(33”)
= 24° + 104’ + 78” + 60° + 96’ + 132”
= 24° + (3 × 30) + 14
Question no – (15)
Solution :
(a) ∠ABC + ∠PQR
= 45°32’29” + 22°45’25”
= 68°17’54”
(b) ∠ABC – ∠PQR
= 45°32’29” – 22°45’25”
= 22°47’4”
Question no – (16)
Solution :
(a) 2∠x – ∠y
= 2 × (35°42’22”) – (45°15’9”)
= 70°84’44” – 45°15’9”
= 26°9’35”
(b) 5∠y
= 5 × 45°15’9”
= 225° 75’ 45”
= 226° 15’ 45”
Angles Exercise 13.2 Solution :
Question no – (1)
Solution :
Adjacent angles – a, b.
Question no –(2)
Solution :
∠YOZ = (199 – 112)°
= 87°
Question no – (3)
Solution :
(a) Complement
= (90° – 45°)
= 45°
(b) Complement
= (90° – 50°)
= 40
(c) Complement
= (90° – 62°)
= 28°
(d) Complement
= (1/5 × 90°)
= 18°
= (90° – 18)
= 72°
Question no – (4)
Solution :
(a) Supplementary angle
= (180° – 17°)
= 163°
(b) Supplementary angle
= (180° – 57°)
= 123°
(c) Supplementary angle
= (180° – 147°)
= 33°
(d) Supplementary angle
= (180° – 1/4 × 90°)
= (180° – 22.5)
= 157.5°
Question no – (5)
Solution :
(a) 90°
(b) 45°
Question no – (6)
Solution :
Supplement,
= (180 – 102 24’ 35”)
= 77° 35’ 25”
Question no – (7)
Solution :
∴ (x + 7) + (2x + 11) = 90°
=> x + 7 + 2x + 11 = 90°
=> 3x = 90° – 18
=> 72
=> x = 72/3
= 24°
Question no – (8)
Solution :
Let, angle 3x & 7x
∴ 3x + 7x = 180°
=> 10x = 180°
=> x = 18°
∴ 3x = (3 × 18°) = 54°
and, 7x = (7 × 18°)
= 126°
Question no – (9)
Solution :
Let, angles 3x & 5x
∴ 5x + 3x = 136°
=> 8x = 136°
=> x = 136/8
= 17°
∴ 3x = (3 × 17)° = 51°
5x = (5 × 17)° = 85°
Question no – (10)
Solution :
From question,
x + x + 40° = 180°
=> 2x = 180 – 40
= 140
=> x = 70°
and, (x + 40°)
= (70 + 40)
= 110°
Question no – (11)
Solution :
From question,
x – 15 + x + 25 = 180°
=> 2x + 10 = 180°
=> x = 180 – 10/2
= 170/2
= 85°
∴ (x – 15°) = (85 – 15)° = 70°
and, (x + 25°) = (85 + 25) = 110°
Question no – (13)
Solution :
(a) 11x + 37 = 180°
=> 11x = 180 – 37°
=> x = 143/11
= 13°
(b) x + 15° + x + 5° + 50° = 180°
=> 2x + 70° = 180°
=> 2x = 180 – 70
=> x = 110/2
= 55°
Question no – (14)
Solution :
P = 3x, q = 4x, r = 3x
Let,
∴ 3x + 4x + 3x = 120°
=> 10x = 120°
=> x = 12°
∴ p = 36°
q = 48°
r = 36°
Question no – (15)
Solution :
∠AOE = 125°
∴ ∠AOD = 125°/2
and, ∠BOE = (180° – 125°)
= 55°
∴ ∠BOC = 55°/2
Now, ∠COD = (125°/2 – 55°/2)
= 125 – 55/2
= 70°
Question no – (16)
Solution :
Here, ∠BOC = 40°
So, ∠AOD = 40°
∴ ∠BOC + ∠AOD = 80°
Now, (360° – 80°) = 280°
∴ ∠x = 280/2
= 140°
∠z = 140°
∠y = 40°
Next Chapter Solution :
👉 Chapter 14 👈
